表示为邻接表的 n 叉树（无环图）的 DFS

给定一棵由 n 个节点组成的树，我们需要打印它的 DFS。
例子：

Input : Edges of graph
        1 2
        1 3
        2 4
        3 5
Output : 1 2 4 3 5

一个简单的解决方案是执行标准 DFS。
我们可以修改我们的方法以避免访问节点的额外空间。我们可以跟踪父级，而不是使用访问过的数组。我们遍历除父节点之外的所有相邻节点。
下面是实现：

C++

/* CPP code to perform DFS of given tree : */
#include 
using namespace std;
 
// DFS on tree
void dfs(vector list[], int node, int arrival)
{
    // Printing traversed node
    cout << node << '\n';
 
    // Traversing adjacent edges
    for (int i = 0; i < list[node].size(); i++) {
 
        // Not traversing the parent node
        if (list[node][i] != arrival)
            dfs(list, list[node][i], node);
    }
}
 
int main()
{
    // Number of nodes
    int nodes = 5;
 
    // Adjacency list
    vector list[10000];
 
    // Designing the tree
    list[1].push_back(2);
    list[2].push_back(1);
 
    list[1].push_back(3);
    list[3].push_back(1);
 
    list[2].push_back(4);
    list[4].push_back(2);
 
    list[3].push_back(5);
    list[5].push_back(3);
 
    // Function call
    dfs(list, 1, 0);
 
    return 0;
}

Java

//JAVA Code For DFS for a n-ary tree (acyclic graph)
// represented as adjacency list
import java.util.*;
 
class GFG {
     
    // DFS on tree
    public static void dfs(LinkedList list[],
                             int node, int arrival)
    {
        // Printing traversed node
        System.out.println(node);
      
        // Traversing adjacent edges
        for (int i = 0; i < list[node].size(); i++) {
      
            // Not traversing the parent node
            if (list[node].get(i) != arrival)
                dfs(list, list[node].get(i), node);
        }
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
 
        // Number of nodes
        int nodes = 5;
      
        // Adjacency list
        LinkedList list[] = new LinkedList[nodes+1];    
         
        for (int i = 0; i < list.length; i ++){
            list[i] = new LinkedList();
        }
         
        // Designing the tree
        list[1].add(2);
        list[2].add(1);
      
        list[1].add(3);
        list[3].add(1);
      
        list[2].add(4);
        list[4].add(2);
      
        list[3].add(5);
        list[5].add(3);
      
        // Function call
        dfs(list, 1, 0);
         
         
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 code to perform DFS of given tree :
 
# DFS on tree
def dfs(List, node, arrival):
     
    # Printing traversed node
    print(node)
 
    # Traversing adjacent edges
    for i in range(len(List[node])):
 
        # Not traversing the parent node
        if (List[node][i] != arrival):
            dfs(List, List[node][i], node)
 
# Driver Code
if __name__ == '__main__':
 
    # Number of nodes
    nodes = 5
 
    # Adjacency List
    List = [[] for i in range(10000)]
 
    # Designing the tree
    List[1].append(2)
    List[2].append(1)
 
    List[1].append(3)
    List[3].append(1)
 
    List[2].append(4)
    List[4].append(2)
 
    List[3].append(5)
    List[5].append(3)
 
    # Function call
    dfs(List, 1, 0)
 
# This code is contributed by PranchalK

C#

// C# Code For DFS for a n-ary tree (acyclic graph)
// represented as adjacency list
using System;
using System.Collections.Generic;
public class GFG {
     
    // DFS on tree
    public static void dfs(List []list,
                            int node, int arrival)
    {
        // Printing traversed node
        Console.WriteLine(node);
     
        // Traversing adjacent edges
        for (int i = 0; i < list[node].Count; i++) {
     
            // Not traversing the parent node
            if (list[node][i] != arrival)
                dfs(list, list[node][i], node);
        }
    }
     
    /* Driver program to test above function */
    public static void Main(String[] args)
    {
 
        // Number of nodes
        int nodes = 5;
     
        // Adjacency list
        List []list = new List[nodes+1];    
         
        for (int i = 0; i < list.Length; i ++){
            list[i] = new List();
        }
         
        // Designing the tree
        list[1].Add(2);
        list[2].Add(1);
     
        list[1].Add(3);
        list[3].Add(1);
     
        list[2].Add(4);
        list[4].Add(2);
     
        list[3].Add(5);
        list[5].Add(3);
     
        // Function call
        dfs(list, 1, 0);
         
         
    }
}
// This code contributed by Rajput-Ji

Javascript

输出：