非递归深度优先搜索算法-Java 学习之路

144

我正在寻找非二叉树的非递归深度优先搜索算法 . 很感谢任何形式的帮助 .

16 回答

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 .

Update: 正如所指出的， take_first() 删除并返回列表中的第一个元素 .

回复于 2024-05-05T04:46:13+08:00

您将使用stack来保存尚未访问过的节点：

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

回复于 2024-05-05T04:46:13+08:00

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通过兄弟指针存储，则可以将下一个兄弟节点存储为：

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

回复于 2024-05-05T04:46:13+08:00

使用堆栈跟踪节点

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)

}

回复于 2024-05-05T04:46:13+08:00

1

虽然"use a stack"可以作为人为的面试问题的答案，但实际上，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 .

回复于 2024-05-05T04:46:13+08:00

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);
  }
}

回复于 2024-05-05T04:46:13+08:00

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（从根开始）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 .

回复于 2024-05-05T04:46:13+08:00

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 .

以下伪代码可以工作（Java和C的混合可读性）：

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顺序通知--DFS从root开始但最后通知它，这与BFS非常简单，实现起来很简单 .

对于踢，请尝试以下图：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，并且应该访问节点的顺序必须是：w，t，v，s DFS的笨拙实现可能首先通知t，这表示存在错误 . DFS的递归实现总是最后到达 .

回复于 2024-05-05T04:46:13+08:00

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typical non-recursive DFS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 .

回复于 2024-05-05T04:46:13+08:00

你可以使用堆栈 . 我使用Adjacency Matrix实现了图形：

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();
        }
    }
}

回复于 2024-05-05T04:46:13+08:00

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;
}

回复于 2024-05-05T04:46:13+08:00

271

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

刚刚观看了这段视频并推出了实施方案 . 我很容易理解 . 请批评这个 .

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()
}

回复于 2024-05-05T04:46:13+08:00

使用 Stack ，以下是要遵循的步骤：然后按下堆栈上的第一个顶点，

如果可能，访问相邻的未访问顶点，对其进行标记，然后将其推入堆栈 .
如果你不能按照步骤1，那么，如果可能的话，从堆栈中弹出一个顶点 .
如果您无法执行步骤1或步骤2，那么您就完成了 .

这是遵循以上步骤的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;
}

回复于 2024-05-05T04:46:13+08:00

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);
            }
        }

回复于 2024-05-05T04:46:13+08:00

基于@ biziclop答案的伪代码：

仅使用基本结构：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：我使用的是1的数组索引，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

回复于 2024-05-05T04:46:13+08:00

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);
}}

输出：广度优先遍历 - 从顶点2开始：2 0 3 1 4深度优先遍历 - 从顶点2开始：2 3 4 1 0

回复于 2024-05-05T04:46:13+08:00

非递归深度优先搜索算法

16 回答

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