# DFS for a n-ary tree (acyclic graph) represented as adjacency list

A tree consisting of n nodes is given, we need to print its DFS.

Examples :

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

A simple solution is to do implement standard DFS
We can modify our approach to avoid extra space for visited nodes. Instead of using the visited array, we can keep track of parent. We traverse all adjacent nodes but the parent.

Below is the implementation :

## C++

 `/* CPP code to perform DFS of given tree : */` `#include ` `using` `namespace` `std;`   `// DFS on tree` `void` `dfs(vector<``int``> 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<``int``> 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<``int``> []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<``int``> []list = ``new` `List<``int``>[nodes+1];     ` `        `  `        ``for` `(``int` `i = 0; i < list.Length; i ++){ ` `            ``list[i] = ``new` `List<``int``>(); ` `        ``} ` `        `  `        ``// 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

 ``

Output

```1
2
4
3
5```

Time Complexity: O(V+E) where V is the number of nodes and E is the number of edges in the graph.

Auxiliary space: O(10000)

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!