Coder's Cat

Depth-First-Search(DFS) Explained With Visualization


DFS Overview

The Depth First Search(DFS) is the most fundamental search algorithm used to explore the nodes and edges of a graph. It runs with time complexity of O(V+E), where V is the number of nodes, and E is the number of edges in a graph.

DFS is often used as a building block in other algorithms; it can be used to:

DFS Visualization on Maze

The source is at the position of left-up, and the target is the position of right-bottom.

As we can see, DFS explores as far as possible along each branch before backtracking:


The maze is generated by disjoint set.

The recursive implementation

#include <list>
#include <vector>
#include <iostream>
using namespace std;

class Graph {
int V;
list<int> *adj;
bool DFS_rec(int v, int t, vector<bool>& visited);

Graph(int V);
void addEdge(int v, int w);
bool DFS(int v, int t);

Graph::Graph(int V) {
this->V = V;
adj = new list<int>[V];

void Graph::addEdge(int v, int w) {

bool Graph::DFS_rec(int v, int t, vector<bool>& visited) {
visited[v] = true;
cout << v << " ";
if(v == t) return true; // Find a path

for (list<int>::iterator it; it = adj[v].begin(); it != adj[v].end(); ++it) {
if (!visited[*it] && DFS_rec(*it, t, visited))
return true;
return false;

bool Graph::DFS(int v, int t) {
vector<bool> visited(V, false);
return DFS_rec(v, t, visited);

int main() {
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);

cout << "Following is Depth First Traversal (0 -> 3): \n";
if(g.DFS(0, 3)) cout << "\nFind a Path!" << std::endl;
else cout << "\nNo Path!" << std::endl;
return 0;

The iterative implementation

A non-recursive implementation of DFS needs the data-structure of stack.

The worst-case space complexity is O(E).

bool Graph::DFS(int v, int t) {
vector<bool> marked(V, false);

stack<int> S;
marked[v] = true;
while(!S.empty()) {
int n =; S.pop();
cout << n << " ";
if(n == t) //Find a path to target
return true;
for(list<int>::iterator it = adj[n].begin(); it != adj[n].end(); ++it) {
if(!marked[*it]) {
marked[*it] = true;
return false;


Join my Email List for more insights, It's Free!😋