我正在寻找非二叉树的非递归深度优先搜索算法。非常感谢任何帮助。
答案 0 :(得分：278)
DFS：
list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
currentnode = nodes_to_visit.take_first();
nodes_to_visit.prepend( currentnode.children );
//do something
}
BFS：
list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
currentnode = nodes_to_visit.take_first();
nodes_to_visit.append( currentnode.children );
//do something
}
两者的对称非常酷。
更新：正如所指出的，take_first()会删除并返回列表中的第一个元素。
答案 1 :(得分：36)
您将使用包含尚未访问过的节点的stack：
stack.push(root)
while !stack.isEmpty() do
node = stack.pop()
for each node.childNodes do
stack.push(stack)
endfor
// …
endwhile
答案 2 :(得分：30)
如果您有指向父节点的指针，则可以在没有额外内存的情况下执行此操作。
def dfs(root):
node = root
while True:
visit(node)
if node.first_child:
node = node.first_child # walk down
else:
while not node.next_sibling:
if node is root:
return
node = node.parent # walk up ...
node = node.next_sibling # ... and right
请注意，如果子节点存储为数组而不是兄弟指针，则可以找到下一个兄弟节点：
def next_sibling(node):
try:
i = node.parent.child_nodes.index(node)
return node.parent.child_nodes[i+1]
except (IndexError, AttributeError):
return None
答案 3 :(得分：5)
使用堆栈跟踪您的节点
Stack<Node> s;
s.prepend(tree.head);
while(!s.empty) {
Node n = s.poll_front // gets first node
// do something with q?
for each child of n: s.prepend(child)
}
答案 4 :(得分：3)
虽然“使用堆栈”可能可以作为设计面试问题的答案，但事实上，它只是明确地做了一个递归程序在幕后做的事情。
递归使用程序内置堆栈。当你调用一个函数时，它会将函数的参数推送到堆栈上，当函数返回时，它会弹出程序堆栈。
答案 5 :(得分：3)
基于biziclops的ES6实现很棒的答案：
root = {
text: "root",
children: [{
text: "c1",
children: [{
text: "c11"
}, {
text: "c12"
}]
}, {
text: "c2",
children: [{
text: "c21"
}, {
text: "c22"
}]
}, ]
}
console.log("DFS:")
DFS(root, node => node.children, node => console.log(node.text));
console.log("BFS:")
BFS(root, node => node.children, node => console.log(node.text));
function BFS(root, getChildren, visit) {
let nodesToVisit = [root];
while (nodesToVisit.length > 0) {
const currentNode = nodesToVisit.shift();
nodesToVisit = [
...nodesToVisit,
...(getChildren(currentNode) || []),
];
visit(currentNode);
}
}
function DFS(root, getChildren, visit) {
let nodesToVisit = [root];
while (nodesToVisit.length > 0) {
const currentNode = nodesToVisit.shift();
nodesToVisit = [
...(getChildren(currentNode) || []),
...nodesToVisit,
];
visit(currentNode);
}
}
答案 6 :(得分：2)
PreOrderTraversal is same as DFS in binary tree. You can do the same recursion
taking care of Stack as below.
public void IterativePreOrder(Tree root)
{
if (root == null)
return;
Stack s<Tree> = new Stack<Tree>();
s.Push(root);
while (s.Count != 0)
{
Tree b = s.Pop();
Console.Write(b.Data + " ");
if (b.Right != null)
s.Push(b.Right);
if (b.Left != null)
s.Push(b.Left);
}
}
一般逻辑是，将节点（从root开始）推入Stack，Pop（）和Print（）值。然后，如果它有子（左和右）将它们推入堆栈 - 首先向右推，这样您将首先访问左子（在访问节点本身之后）。当stack为空（）时，您将访问预订中的所有节点。
答案 7 :(得分：1)
假设您要在访问图表中的每个节点时执行通知。简单的递归实现是：
void DFSRecursive(Node n, Set<Node> visited) {
visited.add(n);
for (Node x : neighbors_of(n)) { // iterate over all neighbors
if (!visited.contains(x)) {
DFSRecursive(x, visited);
}
}
OnVisit(n); // callback to say node is finally visited, after all its non-visited neighbors
}
好的，现在你想要一个基于堆栈的实现，因为你的例子不起作用。例如，复杂的图形可能会导致程序堆栈崩溃，您需要实现非递归版本。最大的问题是知道何时发出通知。
以下伪代码可用（Java和C ++混合以提高可读性）：
void DFS(Node root) {
Set<Node> visited;
Set<Node> toNotify; // nodes we want to notify
Stack<Node> stack;
stack.add(root);
toNotify.add(root); // we won't pop nodes from this until DFS is done
while (!stack.empty()) {
Node current = stack.pop();
visited.add(current);
for (Node x : neighbors_of(current)) {
if (!visited.contains(x)) {
stack.add(x);
toNotify.add(x);
}
}
}
// Now issue notifications. toNotifyStack might contain duplicates (will never
// happen in a tree but easily happens in a graph)
Set<Node> notified;
while (!toNotify.empty()) {
Node n = toNotify.pop();
if (!toNotify.contains(n)) {
OnVisit(n); // issue callback
toNotify.add(n);
}
}
它看起来很复杂但发布通知所需的额外逻辑是因为你需要以相反的访问顺序通知--DFS从root开始但最后通知它，不像BFS那样实现起来非常简单。
对于踢球，请尝试以下图表： 节点是s，t，v和w。 有向边是： s-> t，s-> v，t-> w，v-> w，和v-> t。 运行您自己的DFS实现，并且访问节点的顺序必须是： w，t，v，s 一个笨拙的DFS实现可能会先通知t，这表示存在错误。 DFS的递归实现总是会持续到最后。
答案 8 :(得分：1)
完整的示例工作代码，不带堆栈：
import java.util.*;
class Graph {
private List<List<Integer>> adj;
Graph(int numOfVertices) {
this.adj = new ArrayList<>();
for (int i = 0; i < numOfVertices; ++i)
adj.add(i, new ArrayList<>());
}
void addEdge(int v, int w) {
adj.get(v).add(w); // Add w to v's list.
}
void DFS(int v) {
int nodesToVisitIndex = 0;
List<Integer> nodesToVisit = new ArrayList<>();
nodesToVisit.add(v);
while (nodesToVisitIndex < nodesToVisit.size()) {
Integer nextChild= nodesToVisit.get(nodesToVisitIndex++);// get the node and mark it as visited node by inc the index over the element.
for (Integer s : adj.get(nextChild)) {
if (!nodesToVisit.contains(s)) {
nodesToVisit.add(nodesToVisitIndex, s);// add the node to the HEAD of the unvisited nodes list.
}
}
System.out.println(nextChild);
}
}
void BFS(int v) {
int nodesToVisitIndex = 0;
List<Integer> nodesToVisit = new ArrayList<>();
nodesToVisit.add(v);
while (nodesToVisitIndex < nodesToVisit.size()) {
Integer nextChild= nodesToVisit.get(nodesToVisitIndex++);// get the node and mark it as visited node by inc the index over the element.
for (Integer s : adj.get(nextChild)) {
if (!nodesToVisit.contains(s)) {
nodesToVisit.add(s);// add the node to the END of the unvisited node list.
}
}
System.out.println(nextChild);
}
}
public static void main(String args[]) {
Graph g = new Graph(5);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
g.addEdge(3, 1);
g.addEdge(3, 4);
System.out.println("Breadth First Traversal- starting from vertex 2:");
g.BFS(2);
System.out.println("Depth First Traversal- starting from vertex 2:");
g.DFS(2);
}}
输出： 广度优先遍历-从顶点2开始 2 0 3 1个 4 深度优先遍历-从顶点2开始 2 3 4 1个 0
答案 9 :(得分：1)
使用ES6生成器的非递归DFS
class Node {
constructor(name, childNodes) {
this.name = name;
this.childNodes = childNodes;
this.visited = false;
}
}
function *dfs(s) {
let stack = [];
stack.push(s);
stackLoop: while (stack.length) {
let u = stack[stack.length - 1]; // peek
if (!u.visited) {
u.visited = true; // grey - visited
yield u;
}
for (let v of u.childNodes) {
if (!v.visited) {
stack.push(v);
continue stackLoop;
}
}
stack.pop(); // black - all reachable descendants were processed
}
}
它与typical non-recursive DFS不同，可以轻松检测给定节点的所有可到达后代何时被处理，并维护列表/堆栈中的当前路径。
答案 10 :(得分：0)
只想将我的python实现添加到一长串解决方案中。这种非递归算法具有发现事件和完成事件。
worklist = [root_node]
visited = set()
while worklist:
node = worklist[-1]
if node in visited:
# Node is finished
worklist.pop()
else:
# Node is discovered
visited.add(node)
for child in node.children:
worklist.append(child)
答案 11 :(得分：0)
这里是指向Java程序的链接，该Java程序同时显示了递归和非递归方法的DFS，并且还计算了 发现 和 完成 时间，但没有边缘间隙。
public void DFSIterative() {
Reset();
Stack<Vertex> s = new Stack<>();
for (Vertex v : vertices.values()) {
if (!v.visited) {
v.d = ++time;
v.visited = true;
s.push(v);
while (!s.isEmpty()) {
Vertex u = s.peek();
s.pop();
boolean bFinished = true;
for (Vertex w : u.adj) {
if (!w.visited) {
w.visited = true;
w.d = ++time;
w.p = u;
s.push(w);
bFinished = false;
break;
}
}
if (bFinished) {
u.f = ++time;
if (u.p != null)
s.push(u.p);
}
}
}
}
}
完整来源here。
答案 12 :(得分：0)
基于@biziclop答案的伪代码：
getNode(id)和getChildren(id) N 注意：我使用的数组索引从1开始，而不是0。
宽度优先
S = Array(N)
S[1] = 1; // root id
cur = 1;
last = 1
while cur <= last
id = S[cur]
node = getNode(id)
children = getChildren(id)
n = length(children)
for i = 1..n
S[ last+i ] = children[i]
end
last = last+n
cur = cur+1
visit(node)
end
深度优先
S = Array(N)
S[1] = 1; // root id
cur = 1;
while cur > 0
id = S[cur]
node = getNode(id)
children = getChildren(id)
n = length(children)
for i = 1..n
// assuming children are given left-to-right
S[ cur+i-1 ] = children[ n-i+1 ]
// otherwise
// S[ cur+i-1 ] = children[i]
end
cur = cur+n-1
visit(node)
end
答案 13 :(得分：0)
Stack<Node> stack = new Stack<>();
stack.add(root);
while (!stack.isEmpty()) {
Node node = stack.pop();
System.out.print(node.getData() + " ");
Node right = node.getRight();
if (right != null) {
stack.push(right);
}
Node left = node.getLeft();
if (left != null) {
stack.push(left);
}
}
答案 14 :(得分：0)
使用Stack，以下是要遵循的步骤：然后推送堆栈上的第一个顶点，
按照上述步骤执行Java程序：
public void searchDepthFirst() {
// begin at vertex 0
vertexList[0].wasVisited = true;
displayVertex(0);
stack.push(0);
while (!stack.isEmpty()) {
int adjacentVertex = getAdjacentUnvisitedVertex(stack.peek());
// if no such vertex
if (adjacentVertex == -1) {
stack.pop();
} else {
vertexList[adjacentVertex].wasVisited = true;
// Do something
stack.push(adjacentVertex);
}
}
// stack is empty, so we're done, reset flags
for (int j = 0; j < nVerts; j++)
vertexList[j].wasVisited = false;
}
答案 15 :(得分：0)
Java中的DFS迭代：
//DFS: Iterative
private Boolean DFSIterative(Node root, int target) {
if (root == null)
return false;
Stack<Node> _stack = new Stack<Node>();
_stack.push(root);
while (_stack.size() > 0) {
Node temp = _stack.peek();
if (temp.data == target)
return true;
if (temp.left != null)
_stack.push(temp.left);
else if (temp.right != null)
_stack.push(temp.right);
else
_stack.pop();
}
return false;
}
答案 16 :(得分：0)
您可以使用堆栈。我使用Adjacency Matrix实现了图表：
void DFS(int current){
for(int i=1; i<N; i++) visit_table[i]=false;
myStack.push(current);
cout << current << " ";
while(!myStack.empty()){
current = myStack.top();
for(int i=0; i<N; i++){
if(AdjMatrix[current][i] == 1){
if(visit_table[i] == false){
myStack.push(i);
visit_table[i] = true;
cout << i << " ";
}
break;
}
else if(!myStack.empty())
myStack.pop();
}
}
}
答案 17 :(得分：0)
http://www.youtube.com/watch?v=zLZhSSXAwxI
刚观看了这段视频并推出了实施方案。我很容易理解。请批评这一点。
visited_node={root}
stack.push(root)
while(!stack.empty){
unvisited_node = get_unvisited_adj_nodes(stack.top());
If (unvisited_node!=null){
stack.push(unvisited_node);
visited_node+=unvisited_node;
}
else
stack.pop()
}