Check if a given graph is Bipartite using DFS using C++

A bipartite graph is a graph in which if the graph coloring is possible using two colors only i.e.; vertices in a set are colored with the same color. This is a C++ program to check whether a graph bipartite or not using DFS.

Algorithm

Begin
   An array color[] is used to stores 0 or 1 for every node which
   denotes opposite colors.
   Call function DFS from any node.
   If the node w has not been visited previously, then assign !
   color[v] to color[w] and call DFS again to visit nodes connected
   to w.
   If at any instance, color[u] is equal to !color[v], then the node
   is bipartite.
   Modify the DFS function
End

Example

#include<iostream>
#include <bits/stdc++.h>
using namespace std;
void addEd(vector<int> adj[], int w, int v) //adding edge to the graph
{
   adj[w].push_back(v); //add v to w’s list
   adj[v].push_back(w); //add w to v’s list
}
bool Bipartite(vector<int> adj[], int v,
vector<bool>& visited, vector<int>& color)
{
   for (int w : adj[v]) {
      // if vertex w is not explored before
      if (visited[w] == false) {
         // mark present vertex as visited
         visited[w] = true;
         color[w] = !color[v]; //mark color opposite to its parents
         if (!Bipartite(adj, w, visited, color))
            return false;
      }
      // if two adjacent are colored with same color then the graph is not bipartite
      else if (color[w] == color[v])
      return false;
   }
   return true;
}
int main()
{
   int M = 6;
   vector<int> adj[M + 1];
   // to keep a check on whether a node is discovered or not
   vector<bool> visited(M + 1);
   vector<int> color(M + 1); //to color the vertices of
   the graph with 2 color
   addEd(adj, 3, 2);
   addEd(adj, 1, 4 );
   addEd(adj, 2, 1);
   addEd(adj, 5, 3);
   addEd(adj, 6, 2);
   addEd(adj, 3, 1);
   visited[1] = true;
   color[1] = 0;
   if (Bipartite(adj, 1, visited, color)) {
      cout << "Graph is Bipartite";
   } else {
      cout << "Graph is not Bipartite";
   }
   return 0;
}