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Breadth First Search without using Queue

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  • Difficulty Level : Medium
  • Last Updated : 05 May, 2022
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Breadth-first search is a graph traversal algorithm which traverse a graph or tree level by level. In this article, BFS for a Graph is implemented using Adjacency list without using a Queue.
Examples: 

Input: 

Output: BFS traversal = 2, 0, 3, 1 
Explanation: 
In the following graph, we start traversal from vertex 2. When we come to vertex 0, we look for all adjacent vertices of it. 2 is also an adjacent vertex of 0. If we don’t mark visited vertices, then 2 will be processed again and it will become a non-terminating process. Therefore, a Breadth-First Traversal of the following graph is 2, 0, 3, 1. 
 

Approach: This problem can be solved using simple breadth-first traversal from a given source. The implementation uses adjacency list representation of graphs
Here: 
 

  • STL Vector container is used to store lists of adjacent nodes and queue of nodes needed for BFS traversal. 
     
  • A DP array is used to store the distance of the nodes from the source. Every time we move from a node to another node, the distance increases by 1. If the distance to reach the nodes becomes smaller than the previous distance, we update the value stored in the DP[node].

Below is the implementation of the above approach:

CPP




// C++ implementation to demonstrate
// the above mentioned approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distance
// from the source to other nodes
void BFS(int curr, int N, vector<bool>& vis,
         vector<int>& dp, vector<int>& v,
         vector<vector<int> >& adj)
{
 
    while (curr <= N) {
 
        // Current node
        int node = v[curr - 1];
        cout << node << ", ";
 
        for (int i = 0; i < adj[node].size(); i++) {
 
            // Adjacent node
            int next = adj[node][i];
 
            if ((!vis[next])
                && (dp[next] < dp[node] + 1)) {
 
                // Stores the adjacent node
                v.push_back(next);
 
                // Increases the distance
                dp[next] = dp[node] + 1;
 
                // Mark it as visited
                vis[next] = true;
            }
        }
        curr += 1;
    }
}
 
// Function to print the distance
// from source to other nodes
void bfsTraversal(
    vector<vector<int> >& adj,
    int N, int source)
{
    // Initially mark all nodes as false
    vector<bool> vis(N + 1, false);
 
    // Initialize distance array with 0
    vector<int> dp(N + 1, 0), v;
 
    v.push_back(source);
 
    // Initially mark the starting
    // source as 0 and visited as true
    dp = 0;
    vis = true;
 
    // Call the BFS function
    BFS(1, N, vis, dp, v, adj);
}
 
// Driver code
int main()
{
    // No. of nodes in graph
    int N = 4;
 
    // Creating adjacency list
    // for representing graph
    vector<vector<int> > adj(N + 1);
    adj[0].push_back(1);
    adj[0].push_back(2);
    adj[1].push_back(2);
    adj[2].push_back(0);
    adj[2].push_back(3);
    adj[3].push_back(3);
 
    // Following is BFS Traversal
    // starting from vertex 2
    bfsTraversal(adj, N, 2);
 
    return 0;
}

Python3




# C++ implementation to demonstrate
# the above mentioned approach
from queue import Queue
 
# Function to find the distance
# from the source to other nodes
def BFS(curr, N, vis,          dp,  v, adj):
 
    while (curr <= N) :
 
        # Current node
        node = v[curr - 1]
        print(node,end=  ", ")
 
        for i in range(len(adj[node])) :
 
            # Adjacent node
            next = adj[node][i]
 
            if ((not vis[next]) and (dp[next] < dp[node] + 1)) :
 
                # Stores the adjacent node
                v.append(next)
 
                # Increases the distance
                dp[next] = dp[node] + 1
 
                # Mark it as visited
                vis[next] = True
             
         
        curr += 1
     
 
 
# Function to print the distance
# from source to other nodes
def bfsTraversal(adj,    N, source):
    # Initially mark all nodes as false
    vis=[False]*(N + 1)
 
    # Initialize distance array with 0
    dp=[0]*(N + 1); v=[]
 
    v.append(source)
 
    # Initially mark the starting
    # source as 0 and visited as true
    dp = 0
    vis = True
 
    # Call the BFS function
    BFS(1, N, vis, dp, v, adj)
 
 
# Driver code
if __name__ == '__main__':
    # No. of nodes in graph
    N = 4
 
    # Creating adjacency list
    # for representing graph
    adj=[[] for _ in range(N+1)]
    adj[0].append(1)
    adj[0].append(2)
    adj[1].append(2)
    adj[2].append(0)
    adj[2].append(3)
    adj[3].append(3)
 
    # Following is BFS Traversal
    # starting from vertex 2
    bfsTraversal(adj, N, 2)

Output

2, 0, 3, 1, 

Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V)


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