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C++ Program for DFS Traversal

Hello Everyone!

In this tutorial, we will learn how to implement the DFS Traversal on a Graph, in the C++ programming language.

What is DFS Traversal?

As the name suggests, Depth first search (DFS) algorithm starts with the starting node, and then travers each branch of the graph until we find the leaf node which is a node that has no children. The algorithm, then backtracks towards the most recent nodes that is yet to be completely explored. This process is repeated until all the nodes of the graphs are visited or explored.

The data structure used in DFS is Stack. To learn more about the Stack data structure, we will recommend you to visit, where we have explained these concepts in detail.

For better understanding, refer to the well-commented C++ code given below.


#include <iostream>
using namespace std;

int main()
    cout << "\n\nWelcome to Studytonight :-)\n\n\n";
    cout << " =====  Program to demonstrate the DFS Traversal on a Graph, in CPP  ===== \n\n";

    //variable declaration
    int cost[10][10], i, j, k, n, e, top, v, stk[10], visit[10], visited[10];

    cout << "Enter the number of vertices in the Graph: ";
    cin >> n;
    cout << "\nEnter the number of edges in the Graph : ";
    cin >> e;
    cout << "\nEnter the start and end vertex of the edges: \n";

    for (k = 1; k <= e; k++)
        cin >> i >> j;
        cost[i][j] = 1;

    cout << "\nEnter the initial vertex to start the DFS traversal with: ";
    cin >> v;
    cout << "\nThe DFS traversal on the given graph is : \n";
    cout << v << " ";

    //As we start with the vertex v, marking it visited to avoid visiting again
    visited[v] = 1;

    k = 1;

    //The DFS Traversal Logic
    while (k < n)
        for (j = n; j >= 1; j--)
            if (cost[v][j] != 0 && visited[j] != 1 && visit[j] != 1)
                visit[j] = 1;

                //put all the vertices that are connected to the visited vertex into a stack
                stk[top] = j;

        //output all the connected vertices one at a time
        v = stk[--top];
        cout << v << " ";

        //as v is visited so it is not a valid candidate to visit in future so visit[v]=0 and visited[v]=1
        visit[v] = 0;

        //to mark it visited 
        visited[v] = 1;

        cout << "\n\n\n";

    return 0;


C++ DFS Traversal

We hope that this post helped you develop a better understanding of the concept of DFS Traversal and its implementation in C++. For any query, feel free to reach out to us via the comments section down below.

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About the author:
Nikita Pandey is a talented author and expert in programming languages such as C, C++, and Java. Her writing is informative, engaging, and offers practical insights and tips for programmers at all levels.