Breadth First Search or BFS for a Graph
Last Updated : 21 Apr, 2025
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], [0,1,4], [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]
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:
- Initialization: Enqueue the given source vertex into a queue and mark it as visited.
- 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.
- 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.
C++ #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 + " "); } 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.
C++ #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 + " "); } 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
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.
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Water Jug problem using BFSGiven two empty jugs of m and n litres respectively. The jugs don't have markings to allow measuring smaller quantities. You have to use the jugs to measure d litres of water. The task is to find the minimum number of operations to be performed to obtain d litres of water in one of the jugs. In case
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Word Ladder - Set 2 ( Bi-directional BFS )Given a dictionary, and two words start and target (both of the same length). Find length of the smallest chain from start to target if it exists, such that adjacent words in the chain only differ by one character and each word in the chain is a valid word i.e., it exists in the dictionary. It may b
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Implementing Water Supply Problem using Breadth First SearchGiven N cities that are connected using N-1 roads. Between Cities [i, i+1], there exists an edge for all i from 1 to N-1.The task is to set up a connection for water supply. Set the water supply in one city and water gets transported from it to other cities using road transport. Certain cities are b
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Minimum Cost Path in a directed graph via given set of intermediate nodesGiven a weighted, directed graph G, an array V[] consisting of vertices, the task is to find the Minimum Cost Path passing through all the vertices of the set V, from a given source S to a destination D. Examples: Input: V = {7}, S = 0, D = 6 Output: 11 Explanation: Minimum path 0->7->5->6.
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Shortest path in a Binary MazeGiven an M x N matrix where each element can either be 0 or 1. We need to find the shortest path between a given source cell to a destination cell. The path can only be created out of a cell if its value is 1.Note: You can move into an adjacent cell in one of the four directions, Up, Down, Left, and
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Minimum cost to traverse from one index to another in the StringGiven a string S of length N consisting of lower case character, the task is to find the minimum cost to reach from index i to index j. At any index k, the cost to jump to the index k+1 and k-1(without going out of bounds) is 1. Additionally, the cost to jump to any index m such that S[m] = S[k] is
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