假设有一棵树:
1
/ \
2 3
/ \
4 5
然后镜像将是:
1
/ \
3 2
/ \
5 4
假设节点具有以下结构:
struct node{
node left;
node right;
int value;
}
有人可以为此建议算法吗?
答案 0 :(得分:35)
听起来像是家庭作业。
看起来很容易。编写一个递归例程,深度优先访问每个节点,并构建左右反转的镜像树。
struct node *mirror(struct node *here) {
if (here == NULL)
return NULL;
else {
struct node *newNode = malloc (sizeof(struct node));
newNode->value = here->value;
newNode->left = mirror(here->right);
newNode->right = mirror(here->left);
return newNode;
}
}
这将返回一个新树 - 其他一些答案就是这样做的。取决于你的任务要求你做什么:)
答案 1 :(得分:26)
void swap_node(node n) {
if(n != null) {
node tmp = n.left;
n.left = n.right;
n.right = tmp;
swap_node(n.left);
swap_node(n.right);
}
}
swap_node(root);
答案 2 :(得分:10)
Banal解决方案:
for each node in tree
exchange leftchild with rightchild.
答案 3 :(得分:5)
JAVA中的递归和迭代方法: 1)递归:
public static TreeNode mirrorBinaryTree(TreeNode root){
if(root == null || (root.left == null && root.right == null))
return root;
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
mirrorBinaryTree(root.left);
mirrorBinaryTree(root.right);
return root;
}
2)迭代:
public static TreeNode mirrorBinaryTreeIterative(TreeNode root){
if(root == null || (root.left == null && root.right == null))
return root;
TreeNode parent = root;
Stack<TreeNode> treeStack = new Stack<TreeNode>();
treeStack.push(root);
while(!treeStack.empty()){
parent = treeStack.pop();
TreeNode temp = parent.right;
parent.right = parent.left;
parent.left = temp;
if(parent.right != null)
treeStack.push(parent.right);
if(parent.left != null)
treeStack.push(parent.left);
}
return root;
}
答案 4 :(得分:3)
void mirror(struct node* node)
{
if (node==NULL)
{
return;
}
else
{
struct node* temp;
mirror(node->left);
mirror(node->right);
temp = node->left;
node->left = node->right;
node->right = temp;
}
}
答案 5 :(得分:3)
迭代解决方案:
public void mirrorIterative() {
Queue<TreeNode> nodeQ = new LinkedList<TreeNode>();
nodeQ.add(root);
while(!nodeQ.isEmpty()) {
TreeNode node = nodeQ.remove();
if(node.leftChild == null && node.rightChild == null)
continue;
if(node.leftChild != null && node.rightChild != null) {
TreeNode temp = node.leftChild;
node.leftChild = node.rightChild;
node.rightChild = temp;
nodeQ.add(node.leftChild);
nodeQ.add(node.rightChild);
}
else if(node.leftChild == null) {
node.leftChild = node.rightChild;
node.rightChild = null;
nodeQ.add(node.leftChild);
} else {
node.rightChild = node.leftChild;
node.leftChild = null;
nodeQ.add(node.rightChild);
}
}
}
答案 6 :(得分:2)
void mirror(node<t> *& root2,node<t> * root)
{
if(root==NULL)
{
root2=NULL;
}
else {
root2=new node<t>;
root2->data=root->data;
root2->left=NULL;
root2->right=NULL;
mirror(root2->left,root->right);
mirror(root2->right,root->left);
}
}
答案 7 :(得分:1)
TreeNode * mirror(TreeNode *node){
if(node==NULL){
return NULL;
}else{
TreeNode *temp=node->left;
node->left=mirror(node->right);
node->right=mirror(temp);
return node;
}
}
答案 8 :(得分:0)
这是我的功能。建议是否有更好的解决方案:
void mirrorimage(struct node *p)
{
struct node *q;
if(p!=NULL)
{
q=swaptrs(&p);
p=q;
mirrorimage(p->left);
mirrorimage(p->right);
}
}
struct node* swaptrs(struct node **p)
{
struct node *temp;
temp=(*p)->left;
(*p)->left=(*p)->right;
(*p)->right=temp;
return (*p);
}
答案 9 :(得分:0)
递归Java代码
public class TreeMirrorImageCreator {
public static Node createMirrorImage(Node originalNode,Node mirroredNode){
mirroredNode.setValue(originalNode.getValue());
if(originalNode.getLeft() != null){
mirroredNode.setLeft(createMirrorImage(originalNode.getRight(),new Node(0)));
}
if(originalNode.getRight() != null){
mirroredNode.setRight(createMirrorImage(originalNode.getLeft(), new Node(0)));
}
return mirroredNode;
}
}
答案 10 :(得分:0)
struct node *MirrorOfBinaryTree( struct node *root)
{ struct node *temp;
if(root)
{
MirrorOfBinaryTree(root->left);
MirrorOfBinaryTree(root->right);
/*swap the pointers in this node*/
temp=root->right;
root->right=root->left;;
root->left=temp;
}
return root;
}
时间复杂度:O(n) 空间复杂度:O(n)
答案 11 :(得分:0)
嗯,这个问题得到了很多答案。我发布的迭代版本很容易理解。这使用级别顺序遍历。
Call Text_bold(Application.Union(Range("B5:C5"), Range("B6:C6")))
答案 12 :(得分:0)
这是在Python中使用队列的非递归方法。 Queue类可以这样初始化:
String s1="aeiou";
String s2="This is a test string which could be any text";
s2 = string.Concat(s2.Where(c => !s1.Contains(c)));
Console.WriteLine(s2);
代表节点的类:
class Queue(object):
def __init__(self):
self.items = []
def enqueue(self, item):
self.items.insert(0, item)
def dequeue(self):
if not self.is_empty():
return self.items.pop()
def is_empty(self):
return len(self.items) == 0
def peek(self):
if not self.is_empty():
return self.items[-1]
def __len__(self):
return self.size()
def size(self):
return len(self.items)
这是完成主要任务的方法:
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.val = data