非递归深度优先搜索算法

使用堆栈来跟踪节点

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)


}

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
}

石:

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
}

两者的对称相当酷。

正如所指出的，take_first()删除并返回列表中的第一个元素。

你可以使用堆栈来保存尚未访问的节点:

stack.push(root)
while !stack.isEmpty() do
node = stack.pop()
for each node.childNodes do
stack.push(stack)
endfor
// …
endwhile

如果你有指向父节点的指针，你可以在没有额外内存的情况下完成。

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

注意，如果子节点存储为数组而不是通过兄弟指针，那么下一个兄弟节点可以通过以下方式找到:

def next_sibling(node):
try:
i =    node.parent.child_nodes.index(node)
return node.parent.child_nodes[i+1]
except (IndexError, AttributeError):
return None

http://www.youtube.com/watch?v=zLZhSSXAwxI

刚刚看了这个视频，并提出了实施方案。这对我来说似乎很容易理解。请评论一下。

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()
}

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);


}
}

一般的逻辑是，将一个节点(从根开始)推入Stack, Pop()它和Print()值。然后，如果它有子节点(左和右)，将它们推入堆栈-先推右，这样你就会先访问左子节点(在访问节点本身之后)。当stack为空()时，您将访问Pre-Order中的所有节点。

你可以使用堆栈。我用邻接矩阵实现了图:

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();
}
}
}

虽然“使用堆栈”可能可以作为人为的面试问题的答案，但实际上，它只是显式地做递归程序在幕后所做的事情。

递归使用程序内置堆栈。当你调用一个函数时，它将函数的参数推入堆栈，当函数返回时，它通过弹出程序堆栈来执行。

假设您希望在访问图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，现在你需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的堆栈崩溃，您需要实现一个非递归版本。最大的问题是知道何时发出通知。

下面的伪代码可以工作(为了可读性，Java和c++混合使用):

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，但发出通知所需的额外逻辑存在，因为您需要以相反的访问顺序通知- DFS从根开始，但在最后通知它，不像BFS实现非常简单。

为踢，尝试以下图表: 节点是s t v w。 有向边为: S ->t, S ->v, t->w, v->w, v->t。 运行你自己的DFS实现，访问节点的顺序必须是: W t v s 一个笨拙的DFS实现可能会首先通知t，这表明存在错误。DFS的递归实现总是在最后到达w .

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;
}

使用Stack，以下是要遵循的步骤:

1. 如果可能，访问一个相邻的未访问顶点，标记它，
2. 如果你不能遵循第1步，那么，如果可能的话，弹出一个顶点 李堆栈。< / >
3. 如果你不能遵循第1步或第2步，你就完了。

下面是执行上述步骤的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;
}

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
}
}

它偏离典型的非递归DFS，以方便检测给定节点的所有可达后代是否被处理，并维护列表/堆栈中的当前路径。

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);
}
}

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);
}
}

伪代码基于@biziclop的答案:

• 只使用基本结构:变量、数组、if、while和for
• 函数getNode(id)和getChildren(id)
• 假设已知节点数N

注意:我从1开始使用数组索引，而不是0。

广度优先

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

完整的示例工作代码，没有堆栈:

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);
}}

这里是一个java程序的链接，显示DFS同时遵循递归和非递归方法，并计算发现< em > < / em >和< em > < / em >完成时间，但没有边对齐。

    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);
}
}
}
}
}

完整源代码在这里。

只是想把我的python实现添加到长长的解决方案列表中。这种非递归算法具有发现和完成事件。

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)