Breadth First Search or BFS for a Graph

Try it on GfG Practice

Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list containing the BFS traversal of the graph.

Examples:

Input: adj = [[1,2], [0,2,3], [0,4], [1,4], [2,3]]

Output: [0, 1, 2, 3, 4]
Explanation: Starting from 0, the BFS traversal will follow these steps:
Visit 0 → Output: [0]
Visit 1 (first neighbor of 0) → Output: [0, 1]
Visit 2 (next neighbor of 0) → Output: [0, 1, 2]
Visit 3 (next neighbor of 1) → Output: [0, 1, 2,3]
Visit 4 (neighbor of 2) → Final Output: [0, 1, 2, 3, 4]

Input: adj = [[1, 2], [0, 2], [0, 1, 3, 4], [2], [2]]
Output: [0, 1, 2, 3, 4]
Explanation: Starting from 0, the BFS traversal proceeds as follows:
Visit 0 → Output: [0]
Visit 1 (the first neighbor of 0) → Output: [0, 1]
Visit 2 (the next neighbor of 0) → Output: [0, 1, 2]
Visit 3 (the first neighbor of 2 that hasn’t been visited yet) → Output: [0, 1, 2, 3]
Visit 4 (the next neighbor of 2) → Final Output: [0, 1, 2, 3, 4]

Input: adj = [[2, 4], [3], [0, 4], [1, 4], [0, 2, 3]]
Output: [0, 2, 4, 3, 1]
Explanation: Starting the BFS from vertex 0:
Visit vertex 0 → Output: [0]
Visit vertex 2 (neighbor of 0) → Output: [0, 2]
Visit vertex 4 (next neighbor of 0) → Output: [0, 2, 4]
Visit vertex 3 (unvisited neighbor of 4) → Output: [0, 2, 4, 3]
Visit vertex 1 (unvisited neighbor of 3) → Final Output: [0, 2, 4, 3, 1]

What is Breadth First Search?

Breadth First Search (BFS) is a fundamental graph traversal algorithm. It begins with a node, then first traverses all its adjacent nodes. Once all adjacent are visited, then their adjacent are traversed.

• BFS is different from DFS in a way that closest vertices are visited before others. We mainly traverse vertices level by level.
• Popular graph algorithms like Dijkstra’s shortest path, Kahn’s Algorithm, and Prim’s algorithm are based on BFS.
• BFS itself can be used to detect cycle in a directed and undirected graph, find shortest path in an unweighted graph and many more problems.

BFS from a Given Source

The algorithm starts from a given source and explores all reachable vertices from the given source. It is similar to the Breadth-First Traversal of a tree. Like tree, we begin with the given source (in tree, we begin with root) and traverse vertices level by level using a queue data structure. The only catch here is that, unlike trees, graphs may contain cycles, so we may come to the same node again. To avoid processing a node more than once, we use a Boolean visited array.

Follow the below given approach:

1. Initialization: Enqueue the given source vertex into a queue and mark it as visited.
2. Exploration: While the queue is not empty:
   • Dequeue a node from the queue and visit it (e.g., print its value).
   • For each unvisited neighbor of the dequeued node:
   • Enqueue the neighbor into the queue.
   • Mark the neighbor as visited.
3. Termination: Repeat step 2 until the queue is empty.

This algorithm ensures that all nodes in the graph are visited in a breadth-first manner, starting from the starting node.

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

// BFS from given source s
vector<int> bfs(vector<vector<int>>& adj)  {
    int V = adj.size();
    
    int s = 0; // source node 
    // create an array to store the traversal
    vector<int> res;

    // Create a queue for BFS
    queue<int> q;  
    
    // Initially mark all the vertices as not visited
    vector<bool> visited(V, false);

    // Mark source node as visited and enqueue it
    visited[s] = true;
    q.push(s);

    // Iterate over the queue
    while (!q.empty()) {
      
        // Dequeue a vertex from queue and store it
        int curr = q.front();
        q.pop();
        res.push_back(curr);

        // Get all adjacent vertices of the dequeued 
        // vertex curr If an adjacent has not been 
        // visited, mark it visited and enqueue it
        for (int x : adj[curr]) {
            if (!visited[x]) {
                visited[x] = true;
                q.push(x);
            }
        }
    }

    return res;
}

int main()  {

    vector<vector<int>> adj = {{1,2}, {0,2,3}, {0,4}, {1,4}, {2,3}};
    vector<int> ans = bfs(adj);
    for(auto i:ans) {
        cout<<i<<" ";
    }
    return 0;
}

// Function to find BFS of Graph from given source s
import java.util.*;

class GfG {

    // BFS from given source s
    static ArrayList<Integer> bfs(
        ArrayList<ArrayList<Integer>> adj) {
        int V = adj.size();
        
        int s = 0; // source node
        // create an array to store the traversal
        ArrayList<Integer> res = new ArrayList<>();
        
        // Create a queue for BFS
        Queue<Integer> q = new LinkedList<>();
        
        // Initially mark all the vertices as not visited
        boolean[] visited = new boolean[V];
        
        // Mark source node as visited and enqueue it
        visited[s] = true;
        q.add(s);
        
        // Iterate over the queue
        while (!q.isEmpty()) {
            
            // Dequeue a vertex from queue and store it
            int curr = q.poll();
            res.add(curr);
            
            // Get all adjacent vertices of the dequeued 
            // vertex curr If an adjacent has not been 
            // visited, mark it visited and enqueue it
            for (int x : adj.get(curr)) {
                if (!visited[x]) {
                    visited[x] = true;
                    q.add(x);
                }
            }
        }
        return res;
    }
    
    public static void main(String[] args) {
        
        // create the adjacency list
        // { {2, 3, 1}, {0}, {0, 4}, {0}, {2} }
       
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        adj.add(new ArrayList<>(Arrays.asList(1, 2)));
        adj.add(new ArrayList<>(Arrays.asList(0, 2, 3)));       
        adj.add(new ArrayList<>(Arrays.asList(0, 4)));       
        adj.add(new ArrayList<>(Arrays.asList(1,4)));          
        adj.add(new ArrayList<>(Arrays.asList(2,3)));          
        
        
        ArrayList<Integer> ans = bfs(adj);
        for (int i : ans) {
            System.out.print(i + " ");
        }
    }
}

# Function to find BFS of Graph from given source s
def bfs(adj):
    
    # get number of vertices
    V = len(adj)
    
    # create an array to store the traversal
    res = []
    s = 0
    # Create a queue for BFS
    from collections import deque
    q = deque()
    
    # Initially mark all the vertices as not visited
    visited = [False] * V
    
    # Mark source node as visited and enqueue it
    visited[s] = True
    q.append(s)
    
    # Iterate over the queue
    while q:
        
        # Dequeue a vertex from queue and store it
        curr = q.popleft()
        res.append(curr)
        
        # Get all adjacent vertices of the dequeued 
        # vertex curr If an adjacent has not been 
        # visited, mark it visited and enqueue it
        for x in adj[curr]:
            if not visited[x]:
                visited[x] = True
                q.append(x)
                
    return res

if __name__ == "__main__":
    
    # create the adjacency list
    # [ [2, 3, 1], [0], [0, 4], [0], [2] ]
    adj = [[1,2], [0,2,3], [0,4], [1,4], [2,3]]
    ans = bfs(adj)
    for i in ans:
        print(i, end=" ")

// Function to find BFS of Graph from given source s
using System;
using System.Collections.Generic;

class GfG {

    // BFS from given source s
    static List<int> bfs(List<int>[] adj) {
        int V = adj.Length;
        int s = 0; // source node
        // create an array to store the traversal
        List<int> res = new List<int>();
        
        // Create a queue for BFS
        Queue<int> q = new Queue<int>();
        
        // Initially mark all the vertices as not visited
        bool[] visited = new bool[V];
        
        // Mark source node as visited and enqueue it
        visited[s] = true;
        q.Enqueue(s);
        
        // Iterate over the queue
        while (q.Count > 0) {
            
            // Dequeue a vertex from queue and store it
            int curr = q.Dequeue();
            res.Add(curr);
            
            // Get all adjacent vertices of the dequeued 
            // vertex curr If an adjacent has not been 
            // visited, mark it visited and enqueue it
            foreach (int x in adj[curr]) {
                if (!visited[x]) {
                    visited[x] = true;
                    q.Enqueue(x);
                }
            }
        }
        return res;
    }
    
    static void Main() {
        // create the adjacency list
        // { {2, 3, 1}, {0}, {0, 4}, {0}, {2} }
       
        List<int>[] adj = new List<int>[5];
        adj[0] = new List<int> { 1, 2 };
        adj[1] = new List<int> { 0, 2, 3 };
        adj[2] = new List<int> { 0, 4 };
        adj[3] = new List<int> { 1, 4 };
        adj[4] = new List<int> { 2, 3 };
        
        List<int> ans = bfs(adj);
        foreach (int i in ans) {
            Console.Write(i + " ");
        }
    }
}

// Function to find BFS of Graph from given source s
function bfs(adj) {
    let V = adj.length;
    let s = 0; // source node is 0
    // create an array to store the traversal
    let res = [];
    
    // Create a queue for BFS
    let q = [];
    
    // Initially mark all the vertices as not visited
    let visited = new Array(V).fill(false);
    
    // Mark source node as visited and enqueue it
    visited[s] = true;
    q.push(s);
    
    // Iterate over the queue
    while (q.length > 0) {
        
        // Dequeue a vertex from queue and store it
        let curr = q.shift();
        res.push(curr);
        
        // Get all adjacent vertices of the dequeued 
        // vertex curr If an adjacent has not been 
        // visited, mark it visited and enqueue it
        for (let x of adj[curr]) {
            if (!visited[x]) {
                visited[x] = true;
                q.push(x);
            }
        }
    }
    return res;
}
 
// Main execution
let adj = 
    [ [1,2], [0,2,3], [0,4], [1,4], [2,3]];
let src = 0;
let ans = bfs(adj);
for (let i of ans) {
    process.stdout.write(i + " ");
}

0 1 2 3 4 

BFS of the Disconnected Graph

The above implementation takes a source as an input and prints only those vertices that are reachable from the source and would not print all vertices in case of disconnected graph. Let us see the algorithm that prints all vertices without any source and the graph maybe disconnected.

The algorithm is simple, instead of calling BFS for a single vertex, we call the above implemented BFS for all not yet visited vertices one by one.

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

// BFS from given source s
void bfs(vector<vector<int>>& adj, int s, 
        vector<bool>& visited, vector<int> &res) {

    // Create a queue for BFS
    queue<int> q; 

    // Mark source node as visited and enqueue it
    visited[s] = true;
    q.push(s);

    // Iterate over the queue
    while (!q.empty()) {

        // Dequeue a vertex and store it
        int curr = q.front(); 
        q.pop();
        res.push_back(curr);

        // Get all adjacent vertices of the dequeued 
        // vertex curr If an adjacent has not been 
        // visited, mark it visited and enqueue it
        for (int x : adj[curr]) {
            if (!visited[x]) {
                visited[x] = true;
                q.push(x);
            }
        }
    }
}
                      
// Perform BFS for the entire graph which maybe
// disconnected
vector<int> bfsDisconnected(vector<vector<int>>& adj) {
    int V = adj.size();

    // create an array to store the traversal
    vector<int> res;

    // Initially mark all the vertices as not visited
    vector<bool> visited(V, false); 

    // perform BFS for each node
    for (int i = 0; i < adj.size(); ++i) {
        if (!visited[i]) {
            bfs(adj, i, visited, res);
        }
    }

    return res;
}

int main()  {

    vector<vector<int>> adj = { {1, 2}, {0}, {0},
                                {4}, {3, 5}, {4}};
    vector<int> ans = bfsDisconnected(adj);
    for(auto i:ans) {
        cout<<i<<" ";
    }
    return 0;
}

// BFS from given source s
import java.util.*;

class GfG {

    // BFS from given source s
    static ArrayList<Integer> 
        bfsOfGraph(ArrayList<ArrayList<Integer>> adj, 
                int s, boolean[] visited, ArrayList<Integer> res) {

        // Create a queue for BFS
        Queue<Integer> q = new LinkedList<>();

        // Mark source node as visited and enqueue it
        visited[s] = true;
        q.add(s);

        // Iterate over the queue
        while (!q.isEmpty()) {

            // Dequeue a vertex and store it
            int curr = q.poll();
            res.add(curr);

            // Get all adjacent vertices of the dequeued 
            // vertex curr If an adjacent has not been 
            // visited, mark it visited and enqueue it
            for (int x : adj.get(curr)) {
                if (!visited[x]) {
                    visited[x] = true;
                    q.add(x);
                }
            }
        }
        return res;
    }

    // Perform BFS for the entire graph which maybe
    // disconnected
    static ArrayList<Integer> bfsDisconnected(
                ArrayList<ArrayList<Integer>> adj) {
        int V = adj.size();

        // create an array to store the traversal
        ArrayList<Integer> res = new ArrayList<>();

        // Initially mark all the vertices as not visited
        boolean[] visited = new boolean[V];

        // perform BFS for each node
        for (int i = 0; i < V; i++) {
            if (!visited[i]) {
                bfsOfGraph(adj, i, visited, res);
            }
        }
        return res;
    }

    public static void main(String[] args) {
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        adj.add(new ArrayList<>(Arrays.asList(1, 2)));
        adj.add(new ArrayList<>(Arrays.asList(0))); 
        adj.add(new ArrayList<>(Arrays.asList(0)));   
        adj.add(new ArrayList<>(Arrays.asList(4)));
        adj.add(new ArrayList<>(Arrays.asList(3, 5)));
        adj.add(new ArrayList<>(Arrays.asList(4)));  

        int src = 0;
        ArrayList<Integer> ans = bfsDisconnected(adj);
        for (int i : ans) {
            System.out.print(i + " ");
        }
    }
}

# BFS from given source s
from collections import deque

def bfsOfGraph(adj, s, visited, res):
    
    # Create a queue for BFS
    q = deque()
    
    # Mark source node as visited and enqueue it
    visited[s] = True
    q.append(s)
    
    # Iterate over the queue
    while q:
        
        # Dequeue a vertex and store it
        curr = q.popleft()
        res.append(curr)
        
        # Get all adjacent vertices of the dequeued 
        # vertex curr If an adjacent has not been 
        # visited, mark it visited and enqueue it
        for x in adj[curr]:
            if not visited[x]:
                visited[x] = True
                q.append(x)
    return res

# Perform BFS for the entire graph which maybe
# disconnected
def bfsDisconnected(adj):
    V = len(adj)
    
    # create an array to store the traversal
    res = []
    
    # Initially mark all the vertices as not visited
    visited = [False] * V
    
    # perform BFS for each node
    for i in range(V):
        if not visited[i]:
            bfsOfGraph(adj, i, visited, res)
    return res

if __name__ == "__main__":
    adj = [[1, 2], [0], [0],
        [4], [3, 5], [4]]
    ans = bfsDisconnected(adj)
    for i in ans:
        print(i, end=" ")

// BFS from given source s
using System;
using System.Collections.Generic;

class GfG {

    // BFS from given source s
    static List<int> bfsOfGraph(List<int>[] adj, 
                int s, bool[] visited, List<int> res) {

        // Create a queue for BFS
        Queue<int> q = new Queue<int>();

        // Mark source node as visited and enqueue it
        visited[s] = true;
        q.Enqueue(s);

        // Iterate over the queue
        while (q.Count > 0) {

            // Dequeue a vertex and store it
            int curr = q.Dequeue();
            res.Add(curr);

            // Get all adjacent vertices of the dequeued 
            // vertex curr If an adjacent has not been 
            // visited, mark it visited and enqueue it
            foreach (int x in adj[curr]) {
                if (!visited[x]) {
                    visited[x] = true;
                    q.Enqueue(x);
                }
            }
        }
        return res;
    }

    // Perform BFS for the entire graph which maybe
    // disconnected
    static List<int> bfsDisconnected(List<int>[] adj) {
        int V = adj.Length;

        // create an array to store the traversal
        List<int> res = new List<int>();

        // Initially mark all the vertices as not visited
        bool[] visited = new bool[V];

        // perform BFS for each node
        for (int i = 0; i < V; i++) {
            if (!visited[i]) {
                bfsOfGraph(adj, i, visited, res);
            }
        }
        return res;
    }

    static void Main() {
        List<int>[] adj = new List<int>[6];
        adj[0] = new List<int> { 1, 2 };
        adj[1] = new List<int> { 0 };
        adj[2] = new List<int> { 0 };
        adj[3] = new List<int> { 4 };
        adj[4] = new List<int> { 3, 5 };
        adj[5] = new List<int> { 4 };

        List<int> ans = bfsDisconnected(adj);
        foreach (int i in ans) {
            Console.Write(i + " ");
        }
    }
}

// BFS from given source s
function bfsOfGraph(adj, s, visited, res) {
    
    // Create a queue for BFS
    let q = [];
    
    // Mark source node as visited and enqueue it
    visited[s] = true;
    q.push(s);
    
    // Iterate over the queue
    while (q.length > 0) {
        
        // Dequeue a vertex and store it
        let curr = q.shift();
        res.push(curr);
        
        // Get all adjacent vertices of the dequeued 
        // vertex curr If an adjacent has not been 
        // visited, mark it visited and enqueue it
        for (let x of adj[curr]) {
            if (!visited[x]) {
                visited[x] = true;
                q.push(x);
            }
        }
    }
    return res;
}
 
// Perform BFS for the entire graph which maybe
// disconnected
function bfsDisconnected(adj) {
    let V = adj.length;
    
    // create an array to store the traversal
    let res = [];
    
    // Initially mark all the vertices as not visited
    let visited = new Array(V).fill(false);
    
    // perform BFS for each node
    for (let i = 0; i < V; i++) {
        if (!visited[i]) {
            bfsOfGraph(adj, i, visited, res);
        }
    }
    return res;
}
 
// Main execution
let adj =
    [[1, 2], [0], [0],
    [4], [3, 5], [4]];
let ans = bfsDisconnected(adj);
for (let i of ans) {
    process.stdout.write(i + " ");
}

0 1 2 3 4 5 

Complexity Analysis of Breadth-First Search (BFS) Algorithm

Time Complexity: O(V + E), BFS explores all the vertices and edges in the graph. In the worst case, it visits every vertex and edge once. Therefore, the time complexity of BFS is O(V + E), where V and E are the number of vertices and edges in the given graph.

Auxiliary Space: O(V), BFS uses a queue to keep track of the vertices that need to be visited. In the worst case, the queue can contain all the vertices in the graph. Therefore, the space complexity of BFS is O(V).

Applications of BFS in Graphs

BFS has various applications in graph theory and computer science, including:

• Shortest Path Finding: BFS can be used to find the shortest path between two nodes in an unweighted graph. By keeping track of the parent of each node during the traversal, the shortest path can be reconstructed.
• Cycle Detection: BFS can be used to detect cycles in a graph. If a node is visited twice during the traversal, it indicates the presence of a cycle.
• Connected Components: BFS can be used to identify connected components in a graph. Each connected component is a set of nodes that can be reached from each other.
• Topological Sorting: BFS can be used to perform topological sorting on a directed acyclic graph (DAG). Topological sorting arranges the nodes in a linear order such that for any edge (u, v), u appears before v in the order.
• Level Order Traversal of Binary Trees: BFS can be used to perform a level order traversal of a binary tree. This traversal visits all nodes at the same level before moving to the next level.
• Network Routing: BFS can be used to find the shortest path between two nodes in a network, making it useful for routing data packets in network protocols.

Problems on BFS for a Graph

• Find the level of a given node in an Undirected Graph
• Minimize maximum adjacent difference in a path from top-left to bottom-right
• Minimum jump to the same value or adjacent to reach the end of an Array
• Maximum coin in minimum time by skipping K obstacles along the path in Matrix
• Check if all nodes of the Undirected Graph can be visited from the given Node
• Minimum time to visit all nodes of a given Graph at least once
• Minimize moves to the next greater element to reach the end of the Array
• Shortest path by removing K walls
• Minimum time required to infect all the nodes of the Binary tree
• Check if destination of given Matrix is reachable with required values of cells

FAQs on Breadth First Search (BFS) for a Graph

Question 1: What is BFS and how does it work?

Answer: BFS is a graph traversal algorithm that systematically explores a graph by visiting all the vertices at a given level before moving on to the next level. It starts from a starting vertex, enqueues it into a queue, and marks it as visited. Then, it dequeues a vertex from the queue, visits it, and enqueues all its unvisited neighbors into the queue. This process continues until the queue is empty.

Question 2: What are the applications of BFS?

Answer: BFS has various applications, including finding the shortest path in an unweighted graph, detecting cycles in a graph, topologically sorting a directed acyclic graph (DAG), finding connected components in a graph, and solving puzzles like mazes and Sudoku.

Question 3: What is the time complexity of BFS?

Answer: The time complexity of BFS is O(V + E), where V is the number of vertices and E is the number of edges in the graph.

Question 4: What is the space complexity of BFS?

Answer: The space complexity of BFS is O(V), as it uses a queue to keep track of the vertices that need to be visited.

Question 5: What are the advantages of using BFS?

Answer: BFS is simple to implement and efficient for finding the shortest path in an unweighted graph. It also guarantees that all the vertices in the graph are visited.

Related Articles:

• Recent Articles on BFS
• Depth First Traversal
• Applications of Breadth First Traversal
• Applications of Depth First Search
• Time and Space Complexity of Breadth First Search (BFS)
  

GeeksforGeeks

Improve

Article Tags :

• Data Structures
• DSA
• Graph
• Queue
• BFS
• graph-basics

+2 More

Practice Tags :

• BFS
• Data Structures
• Graph
• Queue