BFS of graph

Problem Statement - link #

Given a directed graph. The task is to do Breadth First Traversal of this graph starting from 0. Note: One can move from node u to node v only if there’s an edge from u to v and find the BFS traversal of the graph starting from the 0th vertex, from left to right according to the graph. Also, you should only take nodes directly or indirectly connected from Node 0 in consideration.

Your Task: You don’t need to read input or print anything. Your task is to complete the function bfsOfGraph() which takes the integer V denoting the number of vertices and adjacency list as input parameters and returns a list containing the BFS traversal of the graph starting from the 0th vertex from left to right..

Expected Time Complexity: O(V+E)
Expected Auxiliary Space: O(V)

Examples #

Example 1:

Input:

Output: 0 1 2 3 4
Explanation: 
0 is connected to 1 , 2 , 3.
2 is connected to 4.
so starting from 0, it will go to 1 then 2
then 3.After this 2 to 4, thus bfs will be
0 1 2 3 4.

Example 2:

Input:
n = 5
arr[] = {4, 2, 7, 6, 9}
Output: 
62 
Explanation:
First, connect ropes 4 and 2, which makes
the array {6,7,6,9}. Next, add ropes 6 and
6, which results in {12,7,9}. Then, add 7
and 9, which makes the array {12,16}. And
finally add these two which gives {28}.
Hence, the total cost is 6 + 12 + 16 + 
28 = 62.

Constraints #

• 1 ≤ V, E ≤ 10^4

Solutions #

class Solution
{
    public:
    // Function to return Breadth First Traversal of given graph.
    vector<int> bfsOfGraph(int V, vector<int> adj[]) {
        // Code here
        queue<int> q; q.push(0);
        bool vis[V]={false};
        vector<int> res;
        while(!q.empty()){
            int u = q.front(); q.pop();
            vis[u]=true;
            res.push_back(u);
            for(int i: adj[u])
                if(!vis[i])
                   { vis[i]=1;q.push(i);}
        }
        return res;
    }
};