DFS for a n-ary tree (acyclic graph) represented as adjacency list
Last Updated : 14 Mar, 2023
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 <bits/stdc++.h> 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 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<Integer> 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<Integer> list[] = new LinkedList[nodes+1]; for (int i = 0; i < list.length; i ++){ list[i] = new LinkedList<Integer>(); } // 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 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# 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 <script> // JavaScript Code For DFS for a n-ary tree (acyclic graph) // represented as adjacency list // DFS on tree function dfs(list, node, arrival) { // Printing traversed node document.write(node + "</br>"); // Traversing adjacent edges for (let i = 0; i < list[node].length; i++) { // Not traversing the parent node if (list[node][i] != arrival) dfs(list, list[node][i], node); } } // Number of nodes let nodes = 5; // Adjacency list let list = new Array(nodes+1); for (let i = 0; i < list.length; i ++){ list[i] = []; } // Designing the tree list[1].push(2); list[2].push(1); list[1].push(3); list[3].push(1); list[2].push(4); list[4].push(2); list[3].push(5); list[5].push(3); // Function call dfs(list, 1, 0); </script> 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)
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