Check if a given graph is Bipartite using DFS

Given a graph with V vertices numbered from 0 to V-1 and a list of edges, determine whether the graph is bipartite or not.

Note: A bipartite graph is a type of graph where the set of vertices can be divided into two disjoint sets, say U and V, such that every edge connects a vertex in U to a vertex in V, there are no edges between vertices within the same set.

Example:

Input: V = 4, edges[][]= [[0, 1], [0, 2], [1, 2], [2, 3]]
Output: false
Explanation:

The graph is not bipartite because no matter how we try to color the nodes using two colors, there exists a cycle of odd length (like 1–2–0–1), which leads to a situation where two adjacent nodes end up with the same color. This violates the bipartite condition, which requires that no two connected nodes share the same color.

Input: V = 4, edges[][] = [[0, 1], [1, 2], [2, 3]]
Output: true
Explanation:

The given graph can be colored in two colors so, it is a bipartite graph.

In the previous post, we have discussed a BFS approach. In this post, we will cover the DFS approach.

Approach: Using Depth-First Search - O(V+E) Time and O(V) Space

To check if a graph is bipartite using Depth-First Search (DFS), we need to color the graph with two colors such that no two adjacent vertices share the same color. We start from any uncolored vertex, assigning it a color (e.g., color 0). As we explore each vertex, we recursively color its uncolored neighbors with the another color. If we ever find a neighbor that shares the same color as the current vertex, we can simply conclude that the graph is not bipartite. If there is no conflict found after the traversal then the given graph is bipartite.

Step-by-step approach:

Create a colors array where all vertices are initialized to -1 (uncolored).
Iterate through all vertices and for each vertex, if it is uncolored, start a DFS from that vertex.
- Assign a color (0 or 1) to the current vertex.
- Visit all adjacent vertices, if its color is same as current vertex, return false (not bipartite), otherwise assign opposite color and continue DFS.
If all vertices can be colored without conflicts, return true (graph is bipartite).

C++

//Driver Code Starts
#include <bits/stdc++.h>
using namespace std;


vector<vector<int>> constructadj(int V, vector<vector<int>> &edges){
    
    vector<vector<int>> adj(V);
    for(auto it:edges){
        adj[it[0]].push_back(it[1]);
        adj[it[1]].push_back(it[0]);
    }
    return adj;
}

//Driver Code Ends

// Helper function for DFS to check bipartite graph
bool dfs(int u, int color, vector<int> &colors,
                           vector<vector<int>> &adj) {

    // Assign color to the current u
    colors[u] = color; 

    // Iterate through all adjacent vertices
    for(auto &v : adj[u]) {
        if(colors[v] == -1) {

            // Assign alternate color to the adjacent u
            if(!dfs(v, 1 - color, colors, adj))
                return false;
        }
        else if(colors[v] == color) {
          
            // If the adjacent u has the same color,
            // it's not bipartite
            return false;
        }
    }
    return true;
}

// Function to check if the graph is Bipartite using DFS
bool isBipartite(int V, vector<vector<int>> &edges) {

    // Initialize all vertices as uncolored
    vector<int> colors(V, -1); 
    
    // create adjacency list
    vector<vector<int>> adj = constructadj(V,edges);

    // Check each component of the graph
    for(int i = 0; i < V; i++) {

        // If the vertex is uncolored
        if(colors[i] == -1) { 

            // If DFS fails, the graph is not bipartite
            if(!dfs(i, 0, colors, adj))
                return false; 
        }
    }

    // All vertices can be colored bipartitely
    return true;
}

//Driver Code Starts

int main() {
    
    int V = 4;
    vector<vector<int>> edges = {{0, 1}, {0, 2}, {1, 2}, {2, 3}};
    if(isBipartite(V, edges))
        cout << "true";
    else
        cout << "false";
    
    return 0;
}

//Driver Code Ends

Java

//Driver Code Starts
import java.util.*;

class GfG {
    
    // Function to construct the adjacency list from edges
    static ArrayList<ArrayList<Integer>> constructadj(int V, int[][] edges) {
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        
        for (int i = 0; i < V; i++) {
            adj.add(new ArrayList<>());
        }

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u); 
        }

        return adj;
    }
//Driver Code Ends

    // Helper function for DFS to check bipartite graph
    static boolean dfs(int u, int color, 
                 int[] colors, ArrayList<ArrayList<Integer>> adj){
      
        // Assign color to the current u
        colors[u] = color;

        // Iterate through all adjacent vertices
        for (int v : adj.get(u)) {
            if (colors[v] == -1) {

                // Assign alternate color to the adjacent u
                if (!dfs(v, 1 - color, colors, adj))
                    return false;
            }
            else if (colors[v] == color) {

                // If the adjacent u has the same color,
                // it's not bipartite
                return false;
            }
        }
        return true;
    }

    // Function to check if the graph is Bipartite using DFS
    static boolean isBipartite(int V, int[][] edges){

        // Initialize all vertices as uncolored
        int[] colors = new int[V];
        Arrays.fill(colors, -1);
        
        // create adjacency list
        ArrayList<ArrayList<Integer>> adj = constructadj(V,edges);
        
        // Check each component of the graph
        for (int i = 0; i < V; i++) {

            // If the vertex is uncolored
            if (colors[i] == -1) {

                // If DFS fails, the graph is not bipartite
                if (!dfs(i, 0, colors, adj))
                    return false;
            }
        }

        // All vertices can be colored bipartitely
        return true;
    }


//Driver Code Starts
    public static void main(String[] args)
    {
        int V = 4;

        // Edges of the graph
        int[][] edges = {{0, 1}, {0, 2}, {1, 2}, {2, 3}};

        // Check if the graph is bipartite
        System.out.println(isBipartite(V, edges));
    }
}
//Driver Code Ends

Python

#Driver Code Starts

# Function to construct the adjacency list from edges
def constructadj(V, edges):
    adj = [[] for _ in range(V)]
    for edge in edges:
        u, v = edge
        adj[u].append(v)
        adj[v].append(u)  
    return adj


#Driver Code Ends

def dfs(u, color, colors, adj):
    # Assign color to the current u
    colors[u] = color

    # Iterate through all adjacent vertices
    for v in adj[u]:
        if colors[v] == -1:
            # Assign alternate color to the adjacent u
            if not dfs(v, 1 - color, colors, adj):
                return False
        elif colors[v] == color:
            # If the adjacent u has the same color, it's not bipartite
            return False
    return True

def isBipartite(V, edges):
    # Initialize all vertices as uncolored
    colors = [-1] * V
    
    # create adjacency list
    adj = constructadj(V,edges)
    
    # Check each component of the graph
    for i in range(V):
        # If the vertex is uncolored
        if colors[i] == -1:
            # If DFS fails, the graph is not bipartite
            if not dfs(i, 0, colors, adj):
                return False

    # All vertices can be colored bipartitely
    return True


#Driver Code Starts
if __name__ == "__main__":
    
    V = 4
    edges = [[0, 1], [0, 2], [1, 2], [2, 3]]

    print("true" if isBipartite(V, edges) else "false")


#Driver Code Ends

//Driver Code Starts
using System;
using System.Collections.Generic;

class GfG {
    
    public static List<List<int>> constructadj(int V, List<List<int>> edges)
    {
        List<List<int>> adj = new List<List<int>>();
        for (int i = 0; i < V; i++)
            adj.Add(new List<int>());

        foreach (var edge in edges)
        {
            int u = edge[0];
            int v = edge[1];
            adj[u].Add(v);
            adj[v].Add(u); 
        }

        return adj;
    }
    
//Driver Code Ends

    // Helper function for DFS to check bipartite graph
    static bool dfs(int u, int color, int[] colors,
                    List<List<int> > adj){

        // Assign color to the current u
        colors[u] = color;

        // Iterate through all adjacent vertices
        foreach(int v in adj[u])
        {
            if (colors[v] == -1) {

                // Assign alternate color to the adjacent u
                if (!dfs(v, 1 - color, colors, adj))
                    return false;
            }
            else if (colors[v] == color) {

                // If the adjacent u has the same color,
                // it's not bipartite
                return false;
            }
        }
        return true;
    }

    // Function to check if the graph is Bipartite using DFS
    static bool IsBipartite(int V, List<List<int>> edges){
        // Initialize all vertices as uncolored
        int[] colors = new int[V];
        Array.Fill(colors, -1);
        
        // create adjacency list
        List<List<int>>   adj = constructadj(V, edges);
        
        // Check each component of the graph
        for (int i = 0; i < V; i++) {
            // If the vertex is uncolored
            if (colors[i] == -1) {
                // If DFS fails, the graph is not bipartite
                if (!dfs(i, 0, colors, adj))
                    return false;
            }
        }

        // All vertices can be colored bipartitely
        return true;
    }

//Driver Code Starts

    public static void Main(){
        
        int V = 4;
        List<List<int>> edges = new List<List<int>> {
            new List<int>{0, 1},
            new List<int>{0, 2},
            new List<int>{1, 2},
            new List<int>{2, 3}
        }; 

        if (IsBipartite(V, edges))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");
        
        }
}
//Driver Code Ends

JavaScript

//Driver Code Starts
// Function to construct adjacency list from edges
function constructadj(V, edges) {
    const adj = Array.from({ length: V }, () => []);
    
    for (const [u, v] of edges) {
        adj[u].push(v);
        adj[v].push(u); // undirected graph
    }

    return adj;
}

//Driver Code Ends

function dfs(u, color, colors, adj){

    // Assign color to the current u
    colors[u] = color;

    // Iterate through all adjacent vertices
    for (let v of adj[u]) {
        if (colors[v] === -1) {
        
            // Assign alternate color to the adjacent u
            if (!dfs(v, 1 - color, colors, adj))
                return false;
        }
        else if (colors[v] === color) {
        
            // If the adjacent u has the same color, it's
            // not bipartite
            return false;
        }
    }
    return true;
}

// Function to check if the graph is Bipartite using DFS
function isBipartite(V, edges){

    // Initialize all vertices as uncolored
    const colors = Array(V).fill(-1);
    
    // create adjacency list
    let adj  = constructadj(V,edges);

    // Check each component of the graph
    for (let i = 0; i < V; i++) {
    
        // If the vertex is uncolored
        if (colors[i] === -1) {
        
            // If DFS fails, the graph is not bipartite
            if (!dfs(i, 0, colors, adj))
                return false;
        }
    }

    // All vertices can be colored bipartitely
    return true;
}


//Driver Code Starts
const V = 4;
const adj = Array.from({length : V}, () => []);

let edges = [[0, 1], [0, 2], [1, 2], [2, 3]];

console.log(isBipartite(V, edges));
//Driver Code Ends

Output

true

Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges. This is because DFS explores each vertex and edge exactly once.
Auxiliary Space: O(V), for color array and recursion call stack, We do not count the adjacency list in auxiliary space as it is necessary for representing the input graph.

Check if a given graph is Bipartite using DFS

Approach: Using Depth-First Search - O(V+E) Time and O(V) Space

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