二阶树的发布顺序遍历而不递归

Question

使用递归对二进制树 WITHOUT 进行后期顺序遍历的算法是什么？

Answer 1

这是具有一个堆栈且没有访问标志的版本：

private void postorder(Node head) {
  if (head == null) {
    return;
  }
  LinkedList<Node> stack = new LinkedList<Node>();
  stack.push(head);

  while (!stack.isEmpty()) {
    Node next = stack.peek();

    boolean finishedSubtrees = (next.right == head || next.left == head);
    boolean isLeaf = (next.left == null && next.right == null);
    if (finishedSubtrees || isLeaf) {
      stack.pop();
      System.out.println(next.value);
      head = next;
    }
    else {
      if (next.right != null) {
        stack.push(next.right);
      }
      if (next.left != null) {
        stack.push(next.left);
      }
    }
  }
}

Answer 2

这是一个链接，它提供了两个不使用任何访问标志的其他解决方案。

https://leetcode.com/problems/binary-tree-postorder-traversal/

由于树中缺少父指针，这显然是基于堆栈的解决方案。 （如果有父指针，我们就不需要堆栈。）

我们首先将根节点推送到堆栈。当堆栈不为空时，我们继续从堆栈顶部推送节点的左子节点。如果左孩子不存在，我们会推动其正确的孩子。如果它是叶子节点，我们处理节点并将其从堆栈中弹出。

我们还使用变量来跟踪先前遍历的节点。目的是确定遍历是否正在下降/上升树，我们也可以知道它是否从左/右上升。

如果我们从左边提升树，我们就不想再将它的左子移到堆栈中，如果它的右子存在，它应该继续向下爬树。如果我们从右边提升树，我们应该处理它并将其从堆栈中弹出。

在这3种情况下，我们将处理节点并将其从堆栈中弹出：

1. 节点是叶节点（没有子节点）
2. 我们只是从左边遍历树，没有正确的孩子。
3. 我们只是从右边遍历树。

Answer 3

以下是来自wikipedia的示例：

nonRecursivePostorder(rootNode)
  nodeStack.push(rootNode)
  while (! nodeStack.empty())
    currNode = nodeStack.peek()
    if ((currNode.left != null) and (currNode.left.visited == false))
      nodeStack.push(currNode.left)
    else 
      if ((currNode.right != null) and (currNode.right.visited == false))
        nodeStack.push(currNode.right)
      else
        print currNode.value
        currNode.visited := true
        nodeStack.pop()

Answer 4

这是我用于迭代，后序遍历的方法。我喜欢这种方法，因为：

1. 每个循环周期只处理一次转换，因此很容易理解。
2. 类似的解决方案适用于有序和预订遍历

代码：

<ul></ul>

说明：

你可以考虑这样的步骤：

1. 尝试LEFT
   • 如果左节点存在：再次尝试LEFT
   • 如果左节点不存在：尝试右键
2. 试试吧
   • 如果存在右节点：从那里尝试LEFT
   • 如果没有权利，那么你就是一片叶子：尝试CURR
3. 尝试CURR
   • 打印当前节点
   • 以下所有节点均已执行（下订单）：尝试UP
4. 尝试
   • 如果node是root，则没有UP，所以EXIT
   • 如果从左侧出来，请尝试右侧
   • 如果从右边开始，请尝试CURR

Answer 5

import java.util.Stack;

public class IterativePostOrderTraversal extends BinaryTree {

    public static void iterativePostOrderTraversal(Node root){
        Node cur = root;
        Node pre = root;
        Stack<Node> s = new Stack<Node>();
        if(root!=null)
            s.push(root);
        System.out.println("sysout"+s.isEmpty());
        while(!s.isEmpty()){
            cur = s.peek();
            if(cur==pre||cur==pre.left ||cur==pre.right){// we are traversing down the tree
                if(cur.left!=null){
                    s.push(cur.left);
                }
                else if(cur.right!=null){
                    s.push(cur.right);
                }
                if(cur.left==null && cur.right==null){
                    System.out.println(s.pop().data);
                }
            }else if(pre==cur.left){// we are traversing up the tree from the left
                if(cur.right!=null){
                    s.push(cur.right);
                }else if(cur.right==null){
                    System.out.println(s.pop().data);
                }
            }else if(pre==cur.right){// we are traversing up the tree from the right
                System.out.println(s.pop().data);
            }
            pre=cur;
        }
    }

    public static void main(String args[]){
        BinaryTree bt = new BinaryTree();
        Node root = bt.generateTree();
        iterativePostOrderTraversal(root);
    }


}

Answer 6

这是C ++中的一个解决方案，它不需要任何存储来保存树中的书籍 相反，它使用两个堆栈。一个用于帮助我们遍历，另一个用于存储节点，因此我们可以对它们进行遍历遍历。

std::stack<Node*> leftStack;
std::stack<Node*> rightStack;

Node* currentNode = m_root;
while( !leftStack.empty() || currentNode != NULL )
{
    if( currentNode )
    {
        leftStack.push( currentNode );
        currentNode = currentNode->m_left;
    }
    else
    {
        currentNode = leftStack.top();
        leftStack.pop();

        rightStack.push( currentNode );
        currentNode = currentNode->m_right;
    }
}

while( !rightStack.empty() )
{
    currentNode = rightStack.top();
    rightStack.pop();

    std::cout << currentNode->m_value;
    std::cout << "\n";
}

Answer 7

//带有标志的java版本

public static <T> void printWithFlag(TreeNode<T> root){
    if(null == root) return;

    Stack<TreeNode<T>> stack = new Stack<TreeNode<T>>();
    stack.add(root);

    while(stack.size() > 0){
        if(stack.peek().isVisit()){
            System.out.print(stack.pop().getValue() + "  ");
        }else{

            TreeNode<T> tempNode = stack.peek();
            if(tempNode.getRight()!=null){
                stack.add(tempNode.getRight());
            }

            if(tempNode.getLeft() != null){
                stack.add(tempNode.getLeft());
            }



            tempNode.setVisit(true);


        }
    }
}

Answer 8

    import java.util.Stack;
   class Practice
{

    public static void main(String arr[])
    {
        Practice prc = new Practice();
        TreeNode node1 = (prc).new TreeNode(1);
        TreeNode node2 = (prc).new TreeNode(2);
        TreeNode node3 = (prc).new TreeNode(3);
        TreeNode node4 = (prc).new TreeNode(4);
        TreeNode node5 = (prc).new TreeNode(5);
        TreeNode node6 = (prc).new TreeNode(6);
        TreeNode node7 = (prc).new TreeNode(7);
        node1.left = node2;
        node1.right = node3;
        node2.left = node4;
        node2.right = node5;
        node3.left = node6;
        node3.right = node7;
        postOrderIteratively(node1);
    }

    public static void postOrderIteratively(TreeNode root)
    {
        Stack<Entry> stack = new Stack<Entry>();
        Practice prc = new Practice();
        stack.push((prc).new Entry(root, false));
        while (!stack.isEmpty())
        {
            Entry entry = stack.pop();
            TreeNode node = entry.node;
            if (entry.flag == false)
            {
                if (node.right == null && node.left == null)
                {
                    System.out.println(node.data);
                } else
                {
                    stack.push((prc).new Entry(node, true));
                    if (node.right != null)
                    {
                        stack.push((prc).new Entry(node.right, false));
                    }
                    if (node.left != null)
                    {
                        stack.push((prc).new Entry(node.left, false));
                    }
                }
            } else
            {
                System.out.println(node.data);
            }
        }

    }

    class TreeNode
    {
        int data;
        int leafCount;
        TreeNode left;
        TreeNode right;

        public TreeNode(int data)
        {
            this.data = data;
        }

        public int getLeafCount()
        {
            return leafCount;
        }

        public void setLeafCount(int leafCount)
        {
            this.leafCount = leafCount;
        }

        public TreeNode getLeft()
        {
            return left;
        }

        public void setLeft(TreeNode left)
        {
            this.left = left;
        }

        public TreeNode getRight()
        {
            return right;
        }

        public void setRight(TreeNode right)
        {
            this.right = right;
        }

        @Override
        public String toString()
        {
            return "" + this.data;
        }
    }

    class Entry
    {
        Entry(TreeNode node, boolean flag)
        {
            this.node = node;
            this.flag = flag;
        }

        TreeNode node;
        boolean flag;

        @Override
        public String toString()
        {
            return node.toString();
        }
    }


}

Answer 9

为了编写这些递归方法的迭代等价物，我们可以首先了解递归方法本身如何在程序堆栈上执行。假设节点没有它们的父指针，我们需要为迭代变体管理我们自己的“堆栈”。

开始的一种方法是查看递归方法并标记调用将“恢复”的位置（新的初始调用，或递归调用返回后）。下面这些标记为“RP 0”、“RP 1”等（“恢复点”）。对于后序遍历的情况：

void post(node *x)  
{  
  /* RP 0 */  
  if(x->lc) post(x->lc);  
  /* RP 1 */  
  if(x->rc) post(x->rc);  
  /* RP 2 */  
  process(x);  
}

它的迭代变体：

void post_i(node *root)  
{  
  node *stack[1000];  
  int top;  
  node *popped;  
 
  stack[0] = root;  
  top = 0;  
  popped = NULL;  
 
#define POPPED_A_CHILD() \  
  (popped && (popped == curr->lc || popped == curr->rc))  
 
  while(top >= 0)  
  {  
    node *curr = stack[top];  
 
    if(!POPPED_A_CHILD())  
    {  
      /* type (x: 0) */  
      if(curr->rc || curr->lc)  
      {  
        if(curr->rc) stack[++top] = curr->rc;  
 
        if(curr->lc) stack[++top] = curr->lc;  
 
        popped = NULL;  
        continue;  
      }  
    }  
 
    /* type (x: 2) */  
    process(curr);  
    top--;  
    popped = curr;  
  }  
}

带有 (x: 0) 和 (x: 2) 的代码注释对应于递归方法中的“RP 0”和“RP 2”恢复点。

通过将 lc 和 rc 指针推到一起，我们不再需要在 post(x) 完成执行时将 post(x->lc) 调用保持在恢复点 1 .也就是说，我们可以直接将节点转移到“RP 2”，绕过“RP 1”。因此，在“RP 1”阶段没有节点保持在堆栈上。

POPPED_A_CHILD 宏帮助我们推导出两个恢复点之一（“RP 0”或“RP 2”）。

您可以阅读有关此方法的详细信息 here（“后序遍历”部分）。

Answer 10

我一直在寻找一个效果良好且易于自定义的代码段。螺纹树不是“简单”。双栈解决方案需要O（n）内存。 tcb 的LeetCode解决方案和解决方案有额外的检查和推送......

这是一个翻译成C的经典算法，对我有用：

void postorder_traversal(TreeNode *p, void (*visit)(TreeNode *))
{
    TreeNode   *stack[40];      // simple C stack, no overflow check
    TreeNode  **sp = stack;
    TreeNode   *last_visited = NULL;

    for (; p != NULL; p = p->left)
        *sp++ = p;

    while (sp != stack) {
        p = sp[-1];
        if (p->right == NULL || p->right == last_visited) {
            visit(p);
            last_visited = p;
            sp--;
        } else {
            for (p = p->right; p != NULL; p = p->left)
                *sp++ = p;
        }
    }
}

恕我直言，这个算法比表现良好且可读的wikipedia.org / Tree_traversal伪代码更容易理解。有关光荣的详细信息，请参阅Knuth第1卷中的二叉树练习答案。

Answer 11

这是一个Python版本::

class Node:
    def __init__(self,data):
        self.data = data
        self.left = None
        self.right = None

def postOrderIterative(root):

    if root is None :
        return

    s1 = []
    s2 = []
    s1.append(root)

    while(len(s1)>0):
        node = s1.pop()
        s2.append(node)

        if(node.left!=None):
            s1.append(node.left)

        if(node.right!=None):
            s1.append(node.right)

    while(len(s2)>0):
        node = s2.pop()
        print(node.data)

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
postOrderIterative(root)

这是输出::

Answer 12

因此，您可以使用一个堆栈进行后订单遍历。

private void PostOrderTraversal(Node pos) {
    Stack<Node> stack = new Stack<Node>();
    do {
        if (pos==null && (pos=stack.peek().right)==null) {
            for (visit(stack.peek()); stack.pop()==(stack.isEmpty()?null:stack.peek().right); visit(stack.peek())) {}
        } else if(pos!=null) {
            stack.push(pos);
            pos=pos.left;
        }
    } while (!stack.isEmpty());
}

Answer 13

不使用递归的遍历逻辑

在// Publications class SecondScreen extends StatelessWidget { @override Widget build(BuildContext context) { return Scaffold( appBar: AppBar( title: Text("Publications"), ), body: Padding( padding: const EdgeInsets.all(10.0), child: new SingleChildScrollView( child: new Text( 'Federation\n\n //Need this to be bold' 'Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.\n\n' '\n\nEditor', // Need this to be bold style: new TextStyle(color: Colors.black), ), ), ), ); } } 中，处理顺序为Postorder traversal。因此，在访问其他部分之前，我们需要先访问左侧部分。对于树的每个节点，我们将尝试尽可能向下遍历树。对于每个当前节点，如果存在正确的子节点，则在root不为NULL / None的情况下，在将当前节点推送之前将其压入堆栈。现在从堆栈中弹出一个节点，并检查该节点的右子节点是否存在。如果存在，则检查它是否与top元素相同。如果它们相同，则表明我们还没有完成正确的部分，因此在处理当前节点之前，我们必须先处理正确的部分，然后弹出顶部元素（正确的子元素）并将当前节点推回堆栈中。每次我们的头都是弹出的元素。如果当前元素与top和head不相同，则我们同时完成了left和right部分，因此现在我们可以处理当前节点了。我们必须重复前面的步骤，直到堆栈为空。

left-right-current

Answer 14

两种无需递归即可执行订单遍历的方法：

1.使用一个访问节点的哈希集和一个堆栈进行回溯：

INSERT INTO table2 (column1, column2, column3)
SELECT column1, column2, column3
FROM table1
WHERE condition;

时间复杂度：O（n）
空间复杂度：O（2n）

2.使用树更改方法：

private void postOrderWithoutRecursion(TreeNode root) {
    if (root == null || root.left == null && root.right == null) {
        return;
    }
    Stack<TreeNode> stack = new Stack<>();
    Set<TreeNode> visited = new HashSet<>();
    while (!stack.empty() || root != null) {
        if (root != null) {
            stack.push(root);
            visited.add(root);
            root = root.left;
        } else {
            root = stack.peek();
            if (root.right == null || visited.contains(root.right)) {
                System.out.print(root.val+" ");
                stack.pop();
                root = null;
            } else {
                root = root.right;
            }

        }
    }
}

时间复杂度：O（n）
空间复杂度：O（n）

TreeNode类：

private void postOrderWithoutRecursionAlteringTree(TreeNode root) {
    if (root == null || root.left == null && root.right == null) {
        return;
    }
    Stack<TreeNode> stack = new Stack<>();
    while (!stack.empty() || root != null) {
        if (root != null) {
            stack.push(root);
            root = root.left;
        } else {
            root = stack.peek();
            if (root.right == null) {
                System.out.print(root.val+" ");
                stack.pop();
                root = null;
            } else {
                TreeNode temp = root.right;
                root.right = null;
                root = temp;
            }
        }
    }
}

Answer 15

这是我需要用Python编写的通用树的简短版本（walker为3行）。当然，也可以用于更有限的二叉树。树是节点和子级列表的元组。它只有一个堆栈。显示了用法示例。

def postorder(tree):
    def do_something(x):  # Your function here
        print(x),
    def walk_helper(root_node, calls_to_perform):
        calls_to_perform.append(partial(do_something, root_node[0]))
        for child in root_node[1]:
            calls_to_perform.append(partial(walk_helper, child, calls_to_perform))
    calls_to_perform = []
    calls_to_perform.append(partial(walk_helper, tree, calls_to_perform))
    while calls_to_perform:
        calls_to_perform.pop()()
postorder(('a', [('b', [('c', []), ('d', [])])]))

  

d   C   b   一个

Answer 16

最简单的解决方案，它看起来不是最佳答案，但很容易理解。而且我相信，如果您了解解决方案，则可以对其进行修改，以寻求最佳解决方案

//使用两个堆栈

public List<Integer> postorderTraversal(TreeNode root){

 Stack<TreeNode> st=new Stack<>();
 Stack<TreeNode> st2=new Stack<>();
 ArrayList<Integer> al = new ArrayList<Integer>(); 

    if(root==null)
        return al;

 st.push(root);  //push the root to 1st stack

 while(!st.isEmpty())
 {
     TreeNode curr=st.pop();

     st2.push(curr);

     if(curr.left!=null)
        st.push(curr.left);
     if(curr.right!=null)
         st.push(curr.right);

 }

while(!st2.isEmpty())
    al.add(st2.pop().val);

//this ArrayList contains the postorder traversal

  return al;  
}

Answer 17

这就是我要进行后期订单迭代器的内容：

    class PostOrderIterator
    implements Iterator<T> {

        private Stack<Node<T>> stack;
        private Node<T> prev;

        PostOrderIterator(Node<T> root) {
            this.stack = new Stack<>();
            recurse(root);
            this.prev = this.stack.peek();
        }

        private void recurse(Node<T> node) {
            if(node == null) {
                return;
            }

            while(node != null) {
                stack.push(node);
                node = node.left;
            }
            recurse(stack.peek().right);
        }

        @Override
        public boolean hasNext() {
            return !stack.isEmpty();
        }

        @Override
        public T next() {
            if(stack.peek().right != this.prev) {
                recurse(stack.peek().right);
            }

            Node<T> next = stack.pop();
            this.prev = next;
            return next.value;
        }
    }

基本上，主要思想是您应该考虑初始化过程如何将要打印的第一项放置在堆栈的顶部，而其余的堆栈则遵循递归会触及的节点。剩下的事情到那时才变得容易得多。

从设计的角度来看，PostOrderIterator是通过树类的某些工厂方法公开为Iterator<T>的内部类。

Answer 18

在后遍历中，他首先访问节点的左子节点，然后是其右子节点，最后是节点本身。 这种树遍历方法类似于图的深度优先搜索（DFS）遍历。

时间复杂度：O（n）

空间复杂度：O（n）

以下是python中后遍历的迭代实现：

class Node:
    def __init__(self, data=None, left=None, right=None):
        self.data  = data
        self.left  = left
        self.right = right

def post_order(node):
    if node is None:
        return []
    stack = []
    nodes = []
    last_node_visited = None
    while stack or node:
        if node:
            stack.append(node)
            node = node.left
        else:
            peek_node = stack[-1]
            if peek_node.right and last_node_visited != peek_node.right:
                node = peek_node.right
            else:
                nodes.append(peek_node.data)
                last_node_visited = stack.pop()
    return nodes

def main():
    '''
    Construct the below binary tree:

            15
           /  \
          /    \
         /      \
        10      20
       /  \    /  \
      8   12  16  25

    '''
    root = Node(15)
    root.left  = Node(10)
    root.right = Node(20)
    root.left.left  = Node(8)
    root.left.right = Node(12)
    root.right.left  = Node(16)
    root.right.right = Node(25)
    print(post_order(root))


if __name__ == '__main__':
    main()

Answer 19

module app.directive {

interface MyDirectiveScope extends ng.IScope {
    myModel: any[];
    change(searchAttribute: string);
}

class MyController {
    static $inject = ['$scope', '$filter'];

    constructor(public $scope: MyDirectiveScope, public $filter: ng.IFilterService) {
        //code

        $scope.change = function (searchAttribute) {
            $scope.myModel = this.$filter('filter')($scope.myModel, searchAttribute); //error : this.$filter turns out to be undefined here
        };
    }
}

class MyDirective implements ng.IDirective {

    static instance(): ng.IDirective {
        return new MyDirective;
    }

    restrict = "E";
    replace = true;
    templateUrl = "myTemplate.html";
    controller = MyController;
    scope = {};
}


angular.module("myModule")
    .directive("myDirective", MyDirective.instance);}

Answer 20

/**
 * This code will ensure holding of chain(links) of nodes from the root to till the level of the tree.
 * The number of extra nodes in the memory (other than tree) is height of the tree.
 * I haven't used java stack instead used this ParentChain. 
 * This parent chain is the link for any node from the top(root node) to till its immediate parent.
 * This code will not require any altering of existing BinaryTree (NO flag/parent on all the nodes).
 *  
 *  while visiting the Node 11; ParentChain will be holding the nodes 9 -> 8 -> 7 -> 1 where (-> is parent)
 *  
 *             1                               
              / \               
             /   \              
            /     \             
           /       \            
          /         \           
         /           \          
        /             \         
       /               \        
       2               7               
      / \             /         
     /   \           /          
    /     \         /           
   /       \       /            
   3       6       8               
  / \             /             
 /   \           /              
 4   5           9               
                / \             
                10 11

 *               
 * @author ksugumar
 *
 */
public class InOrderTraversalIterative {
    public static void main(String[] args) {
        BTNode<String> rt;
        String[] dataArray = {"1","2","3","4",null,null,"5",null,null,"6",null,null,"7","8","9","10",null,null,"11",null,null,null,null};
        rt = BTNode.buildBTWithPreOrder(dataArray, new Counter(0));
        BTDisplay.printTreeNode(rt);
        inOrderTravesal(rt);
    }

public static void postOrderTravesal(BTNode<String> root) {
    ParentChain rootChain = new ParentChain(root);
    rootChain.Parent = new ParentChain(null);

    while (root != null) {

        //Going back to parent
        if(rootChain.leftVisited && rootChain.rightVisited) {
            System.out.println(root.data); //Visit the node.
            ParentChain parentChain = rootChain.Parent;
            rootChain.Parent = null; //Avoid the leak
            rootChain = parentChain;
            root = rootChain.root;
            continue;
        }

        //Traverse Left
        if(!rootChain.leftVisited) {
            rootChain.leftVisited = true;
            if (root.left != null) {
                ParentChain local = new ParentChain(root.left); //It is better to use pool to reuse the instances.
                local.Parent = rootChain;
                rootChain = local;
                root = root.left;
                continue;
            }
        } 

        //Traverse RIGHT
        if(!rootChain.rightVisited) {
            rootChain.rightVisited = true;
            if (root.right != null) {
                ParentChain local = new ParentChain(root.right); //It is better to use pool to reuse the instances.
                local.Parent = rootChain;
                rootChain = local;
                root = root.right;
                continue;
            }
        }
    }
}

class ParentChain {
    BTNode<String> root;
    ParentChain Parent;
    boolean leftVisited = false;
    boolean rightVisited = false;

    public ParentChain(BTNode<String> node) {
        this.root = node; 
    }

    @Override
    public String toString() {
        return root.toString();
    }
}

Answer 21

这是具有两个堆栈的Java实现

public static <T> List<T> iPostOrder(BinaryTreeNode<T> root) {
    if (root == null) {
        return Collections.emptyList();
    }
    List<T> result = new ArrayList<T>();
    Deque<BinaryTreeNode<T>> firstLevel = new LinkedList<BinaryTreeNode<T>>();
    Deque<BinaryTreeNode<T>> secondLevel = new LinkedList<BinaryTreeNode<T>>();
    firstLevel.push(root);
    while (!firstLevel.isEmpty()) {
        BinaryTreeNode<T> node = firstLevel.pop();
        secondLevel.push(node);
        if (node.hasLeftChild()) {
            firstLevel.push(node.getLeft());
        }
        if (node.hasRightChild()) {
            firstLevel.push(node.getRight());
        }
    }
    while (!secondLevel.isEmpty()) {
        result.add(secondLevel.pop().getData());            
    }       
    return result;
}

这是单元测试

@Test
public void iterativePostOrderTest() {
    BinaryTreeNode<Integer> bst = BinaryTreeUtil.<Integer>fromInAndPostOrder(new Integer[]{4,2,5,1,6,3,7}, new Integer[]{4,5,2,6,7,3,1});
    assertThat(BinaryTreeUtil.iPostOrder(bst).toArray(new Integer[0]), equalTo(new Integer[]{4,5,2,6,7,3,1}));

}

Answer 22

1.1创建一个空堆栈

2.1在root不为NULL时执行以下操作

$cat build.sbt
val myPort = 9090

jetty(port = myPort)

2.2从堆栈弹出一个项目并将其设置为root。

a) Push root's right child and then root to stack.

b) Set root as root's left child.

2.3当堆栈不为空时，重复步骤2.1和2.2。

Answer 23

深度优先，后期订单，非递归，无堆栈

当您有父母时：

   node_t
   {
     left,
     right
     parent
   }

   traverse(node_t rootNode)
   {
     bool backthreading = false 
     node_t node = rootNode

     while(node <> 0)

        if (node->left <> 0) and backthreading = false then
               node = node->left

            continue 
        endif

         >>> process node here <<<


        if node->right <> 0 then
            lNode = node->right
            backthreading = false
        else
            node = node->parent

            backthreading = true
        endif
    endwhile

Answer 24

很高兴看到这个问题有这么多精神的方法。非常鼓舞人心！

我遇到了这个主题，搜索了一个简单的迭代解决方案，用于删除二叉树实现中的所有节点。我尝试了其中的一些，我在网上的其他地方尝试过类似的东西，但没有一个真的符合我的喜好。

问题是，我正在开发一个用于特定目的的数据库索引模块（比特币区块链索引），我的数据存储在磁盘上，而不是存储在RAM中。我根据需要交换页面，进行自己的内存管理。它速度较慢，但​​速度足够快，而且存储在磁盘而不是RAM上，我没有宗教信仰来防止浪费空间（硬盘很便宜）。

因此我的二叉树中的节点有父指针。那是（所有）我正在谈论的额外空间。我需要父母，因为为了各种目的，我需要在树中进行上升和下降迭代。

考虑到这一点，我很快写下了一小段关于如何完成的伪代码，即在运行时删除节点的后序遍历。它已经实施和测试，并成为我的解决方案的一部分。而且它也很快。

事情是：当节点有父指针时，它真的非常简单，而且因为我可以将父节点的链接归零到“刚刚离开”的节点。

这是迭代后序删除的伪代码：

Node current = root;
while (current)
{
  if (current.left)       current = current.left;  // Dive down left
  else if (current.right) current = current.right; // Dive down right
  else
  {
    // Node "current" is a leaf, i.e. no left or right child
    Node parent = current.parent; // assuming root.parent == null
    if (parent)
    {
      // Null out the parent's link to the just departing node
      if (parent.left == current) parent.left = null;
      else                        parent.right = null;
    }
    delete current;
    current = parent;
  }
}
root = null;

如果您对编写复杂集合的更理论方法感兴趣（例如我的二叉树，它实际上是一个自平衡的红黑树），那么请查看以下链接：

http://opendatastructures.org/versions/edition-0.1e/ods-java/6_2_BinarySearchTree_Unbala.html http://opendatastructures.org/versions/edition-0.1e/ods-java/9_2_RedBlackTree_Simulated_.html https://www.cs.auckland.ac.nz/software/AlgAnim/red_black.html

快乐编码： - ）

SørenFog http://iprotus.eu/

Answer 25

我没有添加节点类，因为它不是特别相关或任何测试用例，而是将它们作为阅读器的练习等。

void postOrderTraversal(node* root)
{
    if(root == NULL)
        return;

    stack<node*> st;
    st.push(root);

    //store most recent 'visited' node
    node* prev=root;

    while(st.size() > 0)
    {
        node* top = st.top();
        if((top->left == NULL && top->right == NULL))
        {
            prev = top;
            cerr<<top->val<<" ";
            st.pop();
            continue;
        }
        else
        {
            //we can check if we are going back up the tree if the current
            //node has a left or right child that was previously outputted
            if((top->left == prev) || (top->right== prev))
            {
                prev = top;
                cerr<<top->val<<" ";
                st.pop();
                continue;
            }

            if(top->right != NULL)
                st.push(top->right);

            if(top->left != NULL)
                st.push(top->left);
        }
    }
    cerr<<endl;
}

运行时间O（n） - 需要访问所有节点 AND空间O（n） - 对于堆栈，最坏情况树是单行链接列表

Answer 26

带有1个堆栈且没有标志的Python：

def postorderTraversal(self, root):
    ret = []
    if not root:
        return ret
    stack = [root]
    current = None
    while stack:
        previous = current
        current = stack.pop()
        if previous and ((previous is current) or (previous is current.left) or (previous is current.right)):
            ret.append(current.val)
        else:
            stack.append(current)
            if current.right:
                stack.append(current.right)
            if current.left:
                stack.append(current.left)

    return ret

更好的是使用类似的语句，顺序遍历也是如此

def inorderTraversal(self, root):
    ret = []
    if not root:
        return ret
    stack = [root]
    current = None
    while stack:
        previous = current
        current = stack.pop()
        if None == previous or previous.left is current or previous.right is current:
            if current.right:
                stack.append(current.right)
            stack.append(current)
            if current.left:
                stack.append(current.left)
        else:
            ret.append(current.val)

    return ret

Answer 27

这里我在c＃（。net）中粘贴不同的版本以供参考： （对于有序迭代遍历，您可以参考：Help me understand Inorder Traversal without using recursion）

1. wiki（http://en.wikipedia.org/wiki/Post-order%5Ftraversal#Implementations）（优雅）
2. 单个堆栈的另一个版本 （＃1和＃2：基本上使用这样的事实：在访问实际节点之前，在后期遍历中访问了正确的子节点 - 因此，我们只依赖于检查堆栈顶部的右子节点是否确实是最后一个订单遍历节点那些被访问过 - 我在下面的代码片段中添加了评论以获取详细信息）
3. 使用两个堆栈版本（参考：http://www.geeksforgeeks.org/iterative-postorder-traversal/） （更简单：基本上是邮政订单遍历反向只是预订序遍历，只需先调用一个简单的调整，然后先访问左边的节点）
4. 使用访客标志（简单）
5. 单元测试

〜

public string PostOrderIterative_WikiVersion()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                BinaryTreeNode lastPostOrderTraversalNode = null;
                BinaryTreeNode iterativeNode = this._root;
                Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
                while ((stack.Count > 0)//stack is not empty
                    || (iterativeNode != null))
                {
                    if (iterativeNode != null)
                    {
                        stack.Push(iterativeNode);
                        iterativeNode = iterativeNode.Left;
                    }
                    else
                    {
                        var stackTop = stack.Peek();
                        if((stackTop.Right != null)
                            && (stackTop.Right != lastPostOrderTraversalNode))
                        {
                            //i.e. last traversal node is not right element, so right sub tree is not
                            //yet, traversed. so we need to start iterating over right tree 
                            //(note left tree is by default traversed by above case)
                            iterativeNode = stackTop.Right;
                        }
                        else
                        {
                            //if either the iterative node is child node (right and left are null)
                            //or, stackTop's right element is nothing but the last traversal node
                            //(i.e; the element can be popped as the right sub tree have been traversed)
                            var top = stack.Pop();
                            Debug.Assert(top == stackTop);
                            nodes.Add(top.Element);
                            lastPostOrderTraversalNode = top;
                        }
                    }
                }
            }
            return this.ListToString(nodes);
        }

这是一个邮件订单遍历一个堆栈（我的版本）

public string PostOrderIterative()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                BinaryTreeNode lastPostOrderTraversalNode = null;
                BinaryTreeNode iterativeNode = null;
                Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
                stack.Push(this._root);
                while(stack.Count > 0)
                {
                    iterativeNode = stack.Pop();
                    if ((iterativeNode.Left == null)
                        && (iterativeNode.Right == null))
                    {
                        nodes.Add(iterativeNode.Element);
                        lastPostOrderTraversalNode = iterativeNode;
                        //make sure the stack is not empty as we need to peek at the top
                        //for ex, a tree with just root node doesn't have to enter loop
                        //and also node Peek() will throw invalidoperationexception
                        //if it is performed if the stack is empty
                        //so, it handles both of them.
                        while(stack.Count > 0) 
                        {
                            var stackTop = stack.Peek();
                            bool removeTop = false;
                            if ((stackTop.Right != null) &&
                                //i.e. last post order traversal node is nothing but right node of 
                                //stacktop. so, all the elements in the right subtree have been visted
                                //So, we can pop the top element
                                (stackTop.Right == lastPostOrderTraversalNode))
                            {
                                //in other words, we can pop the top if whole right subtree is
                                //traversed. i.e. last traversal node should be the right node
                                //as the right node will be traverse once all the subtrees of
                                //right node has been traversed
                                removeTop = true;
                            }
                            else if(
                                //right subtree is null
                                (stackTop.Right == null) 
                                && (stackTop.Left != null) 
                                //last traversal node is nothing but the root of left sub tree node
                                && (stackTop.Left == lastPostOrderTraversalNode))
                            {
                                //in other words, we can pop the top of stack if right subtree is null,
                                //and whole left subtree has been traversed
                                removeTop = true;
                            }
                            else
                            {
                                break;
                            }
                            if(removeTop)
                            {
                                var top = stack.Pop();
                                Debug.Assert(stackTop == top);
                                lastPostOrderTraversalNode = top;
                                nodes.Add(top.Element);
                            }
                        }
                    }
                    else 
                    {
                        stack.Push(iterativeNode);
                        if(iterativeNode.Right != null)
                        {
                            stack.Push(iterativeNode.Right);
                        }
                        if(iterativeNode.Left != null)
                        {
                            stack.Push(iterativeNode.Left);
                        }
                    }
                }
            }
            return this.ListToString(nodes);
        }

使用两个堆栈

public string PostOrderIterative_TwoStacksVersion()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                Stack<BinaryTreeNode> postOrderStack = new Stack<BinaryTreeNode>();
                Stack<BinaryTreeNode> rightLeftPreOrderStack = new Stack<BinaryTreeNode>();
                rightLeftPreOrderStack.Push(this._root);
                while(rightLeftPreOrderStack.Count > 0)
                {
                    var top = rightLeftPreOrderStack.Pop();
                    postOrderStack.Push(top);
                    if(top.Left != null)
                    {
                        rightLeftPreOrderStack.Push(top.Left);
                    }
                    if(top.Right != null)
                    {
                        rightLeftPreOrderStack.Push(top.Right);
                    }
                }
                while(postOrderStack.Count > 0)
                {
                    var top = postOrderStack.Pop();
                    nodes.Add(top.Element);
                }
            }
            return this.ListToString(nodes);
        }

使用C＃（。net）中的访问过的标记：

public string PostOrderIterative()
        {
            List<int> nodes = new List<int>();
            if (null != this._root)
            {
                BinaryTreeNode iterativeNode = null;
                Stack<BinaryTreeNode> stack = new Stack<BinaryTreeNode>();
                stack.Push(this._root);
                while(stack.Count > 0)
                {
                    iterativeNode = stack.Pop();
                    if(iterativeNode.visted)
                    {
                        //reset the flag, for further traversals
                        iterativeNode.visted = false;
                        nodes.Add(iterativeNode.Element);
                    }
                    else
                    {
                        iterativeNode.visted = true;
                        stack.Push(iterativeNode);
                        if(iterativeNode.Right != null)
                        {
                            stack.Push(iterativeNode.Right);
                        }
                        if(iterativeNode.Left != null)
                        {
                            stack.Push(iterativeNode.Left);
                        }
                    }
                }
            }
            return this.ListToString(nodes);
        }

定义：

class BinaryTreeNode
    {
        public int Element;
        public BinaryTreeNode Left;
        public BinaryTreeNode Right;
        public bool visted;
    }

string ListToString(List<int> list)
        {
            string s = string.Join(", ", list);
            return s;
        }

单元测试

[TestMethod]
        public void PostOrderTests()
        {
            int[] a = { 13, 2, 18, 1, 5, 17, 20, 3, 6, 16, 21, 4, 14, 15, 25, 22, 24 };
            BinarySearchTree bst = new BinarySearchTree();
            foreach (int e in a)
            {
                string s1 = bst.PostOrderRecursive();
                string s2 = bst.PostOrderIterativeWithVistedFlag();
                string s3 = bst.PostOrderIterative();
                string s4 = bst.PostOrderIterative_WikiVersion();
                string s5 = bst.PostOrderIterative_TwoStacksVersion();
                Assert.AreEqual(s1, s2);
                Assert.AreEqual(s2, s3);
                Assert.AreEqual(s3, s4);
                Assert.AreEqual(s4, s5);
                bst.Add(e);
                bst.Delete(e);
                bst.Add(e);
                s1 = bst.PostOrderRecursive();
                s2 = bst.PostOrderIterativeWithVistedFlag();
                s3 = bst.PostOrderIterative();
                s4 = bst.PostOrderIterative_WikiVersion();
                s5 = bst.PostOrderIterative_TwoStacksVersion();
                Assert.AreEqual(s1, s2);
                Assert.AreEqual(s2, s3);
                Assert.AreEqual(s3, s4);
                Assert.AreEqual(s4, s5);
            }
            Debug.WriteLine(string.Format("PostOrderIterative: {0}", bst.PostOrderIterative()));
            Debug.WriteLine(string.Format("PostOrderIterative_WikiVersion: {0}", bst.PostOrderIterative_WikiVersion()));
            Debug.WriteLine(string.Format("PostOrderIterative_TwoStacksVersion: {0}", bst.PostOrderIterative_TwoStacksVersion()));
            Debug.WriteLine(string.Format("PostOrderIterativeWithVistedFlag: {0}", bst.PostOrderIterativeWithVistedFlag()));
            Debug.WriteLine(string.Format("PostOrderRecursive: {0}", bst.PostOrderRecursive()));
        }

Answer 28

请参阅此完整的Java实现。只需复制代码并粘贴到编译器中即可。它会工作正常。

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

class Node
{
    Node left;
    String data;
    Node right;

    Node(Node left, String data, Node right)
    {
        this.left = left;
        this.right = right;
        this.data = data;
    }

    public String getData()
    {
        return data;
    }
}

class Tree
{
    Node node;

    //insert
    public void insert(String data)
    {
        if(node == null)
            node = new Node(null,data,null);
        else
        {
            Queue<Node> q = new LinkedList<Node>();
            q.add(node);

            while(q.peek() != null)
            {
                Node temp = q.remove();
                if(temp.left == null)
                {
                    temp.left = new Node(null,data,null);
                    break;
                }
                else
                {
                    q.add(temp.left);
                }

                if(temp.right == null)
                {
                    temp.right = new Node(null,data,null);
                    break;
                }
                else
                {
                    q.add(temp.right);
                }
            }
        }
    }

    public void postorder(Node node)
    {
        if(node == null)
            return;
        postorder(node.left);
        postorder(node.right);
        System.out.print(node.getData()+" --> ");
    }

    public void iterative(Node node)
    {
        Stack<Node> s = new Stack<Node>();
        while(true)
        {
            while(node != null)
            {
                s.push(node);
                node = node.left;
            }



            if(s.peek().right == null)
            {
                node = s.pop();
                System.out.print(node.getData()+" --> ");
                if(node == s.peek().right)
                {
                    System.out.print(s.peek().getData()+" --> ");
                    s.pop();
                }
            }

            if(s.isEmpty())
                break;

            if(s.peek() != null)
            {
                node = s.peek().right;
            }
            else
            {
                node = null;
            }
        }
    }
}

class Main
{
    public static void main(String[] args) 
    {
        Tree t = new Tree();
        t.insert("A");
        t.insert("B");
        t.insert("C");
        t.insert("D");
        t.insert("E");

        t.postorder(t.node);
        System.out.println();

        t.iterative(t.node);
        System.out.println();
    }
}

Answer 29

void postorder_stack(Node * root){
    stack ms;
    ms.top = -1;
    if(root == NULL) return ;

    Node * temp ;
    push(&ms,root);
    Node * prev = NULL;
    while(!is_empty(ms)){
        temp = peek(ms);
        /* case 1. We are nmoving down the tree. */
        if(prev == NULL || prev->left == temp || prev->right == temp){
             if(temp->left)
                  push(&ms,temp->left);
             else if(temp->right)
                  push(&ms,temp->right);
             else {
                /* If node is leaf node */
                   printf("%d ", temp->value);
                   pop(&ms);
             }
         }
         /* case 2. We are moving up the tree from left child */
         if(temp->left == prev){
              if(temp->right)
                   push(&ms,temp->right);
              else
                   printf("%d ", temp->value);
         }

        /* case 3. We are moving up the tree from right child */
         if(temp->right == prev){
              printf("%d ", temp->value);
              pop(&ms);
         }
         prev = temp;
      }

}

Answer 30

python中的简单直观解决方案。

        while stack:
            node = stack.pop()
            if node:
                if isinstance(node,TreeNode):
                    stack.append(node.val)
                    stack.append(node.right)
                    stack.append(node.left)
                else:
                    post.append(node)
        return post