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Count the number of nodes at given level in a tree using BFS.
  • Difficulty Level : Easy
  • Last Updated : 27 Jan, 2021

Given a tree represented as undirected graph. Count the number of nodes at given level l. It may be assumed that vertex 0 is root of the tree.

Examples: 

Input :   7
          0 1
          0 2
          1 3
          1 4 
          1 5
          2 6
          2
Output :  4

Input : 6
        0 1
        0 2
        1 3
        2 4
        2 5
        2
Output : 3

BFS is a traversing algorithm which start traversing from a selected node (source or starting node) and traverse the graph layer wise thus exploring the neighbour nodes (nodes which are directly connected to source node). Then, move towards the next-level neighbour nodes. 
As the name BFS suggests, traverse the graph breadth wise as follows:
1. First move horizontally and visit all the nodes of the current layer. 
2. Move to the next layer.

In this code, while visiting each node, the level of that node is set with an increment in the level of its parent node i.e., level[child] = level[parent] + 1. This is how the level of each node is determined. The root node lies at level zero in the tree.

Explanation :



     0         Level 0
   /   \ 
  1     2      Level 1
/ |\    |
3 4 5   6      Level 2

Given a tree with 7 nodes and 6 edges in which node 0 lies at 0 level. Level of 1 can be updated as : level[1] = level[0] +1 as 0 is the parent node of 1. Similarly, the level of other nodes can be updated by adding 1 to the level of their parent. 
level[2] = level[0] + 1, i.e level[2] = 0 + 1 = 1. 
level[3] = level[1] + 1, i.e level[3] = 1 + 1 = 2. 
level[4] = level[1] + 1, i.e level[4] = 1 + 1 = 2. 
level[5] = level[1] + 1, i.e level[5] = 1 + 1 = 2. 
level[6] = level[2] + 1, i.e level[6] = 1 + 1 = 2.
Then, count of number of nodes which are at level l(i.e, l=2) is 4 (node:- 3, 4, 5, 6) 

C++




// C++ Program to print
// count of nodes
// at given level.
#include <iostream>
#include <list>
 
using namespace std;
 
// This class represents
// a directed graph
// using adjacency
// list representation
class Graph {
    // No. of vertices
    int V;
 
    // Pointer to an
    // array containing
    // adjacency lists
    list<int>* adj;
 
public:
    // Constructor
    Graph(int V);
 
    // function to add
    // an edge to graph
    void addEdge(int v, int w);
 
    // Returns count of nodes at
    // level l from given source.
    int BFS(int s, int l);
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
void Graph::addEdge(int v, int w)
{
    // Add w to v’s list.
    adj[v].push_back(w);
 
    // Add v to w's list.
    adj[w].push_back(v);
}
 
int Graph::BFS(int s, int l)
{
    // Mark all the vertices
    // as not visited
    bool* visited = new bool[V];
    int level[V];
 
    for (int i = 0; i < V; i++) {
        visited[i] = false;
        level[i] = 0;
    }
 
    // Create a queue for BFS
    list<int> queue;
 
    // Mark the current node as
    // visited and enqueue it
    visited[s] = true;
    queue.push_back(s);
    level[s] = 0;
 
    while (!queue.empty()) {
 
        // Dequeue a vertex from
        // queue and print it
        s = queue.front();
        queue.pop_front();
 
        // Get all adjacent vertices
        // of the dequeued vertex s.
        // If a adjacent has not been
        // visited, then mark it
        // visited and enqueue it
        for (auto i = adj[s].begin();
                  i != adj[s].end(); ++i) {
            if (!visited[*i]) {
 
                // Setting the level
                // of each node with
                // an increment in the
                // level of parent node
                level[*i] = level[s] + 1;
                visited[*i] = true;
                queue.push_back(*i);
            }
        }
    }
 
    int count = 0;
    for (int i = 0; i < V; i++)
        if (level[i] == l)
            count++;   
    return count; 
}
 
// Driver program to test
// methods of graph class
int main()
{
    // Create a graph given
    // in the above diagram
    Graph g(6);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 3);
    g.addEdge(2, 4);
    g.addEdge(2, 5);
 
    int level = 2;
 
    cout << g.BFS(0, level);
 
    return 0;
}


Java




// Java Program to print
// count of nodes
// at given level.
import java.util.*;
 
// This class represents
// a directed graph
// using adjacency
// list representation
class Graph
{
 
  // No. of vertices
  int V;
 
  // Pointer to an
  // array containing
  // adjacency lists
  Vector<Integer>[] adj;
 
  // Constructor
  @SuppressWarnings("unchecked")
  Graph(int V)
  {
    adj = new Vector[V];
    for (int i = 0; i < adj.length; i++)
    {
      adj[i] = new Vector<>();
    }
    this.V = V;
  }
 
  void addEdge(int v, int w)
  {
 
    // Add w to v’s list.
    adj[v].add(w);
 
    // Add v to w's list.
    adj[w].add(v);
  }
 
  int BFS(int s, int l)
  {
 
    // Mark all the vertices
    // as not visited
    boolean[] visited = new boolean[V];
    int[] level = new int[V];
 
    for (int i = 0; i < V; i++)
    {
      visited[i] = false;
      level[i] = 0;
    }
 
    // Create a queue for BFS
    Queue<Integer> queue = new LinkedList<>();
 
    // Mark the current node as
    // visited and enqueue it
    visited[s] = true;
    queue.add(s);
    level[s] = 0;
    int count = 0;
    while (!queue.isEmpty())
    {
 
      // Dequeue a vertex from
      // queue and print it
      s = queue.peek();
      queue.poll();
 
      Vector<Integer> list = adj[s];
      // Get all adjacent vertices
      // of the dequeued vertex s.
      // If a adjacent has not been
      // visited, then mark it
      // visited and enqueue it
      for (int i : list)
      {
        if (!visited[i])
        {
          visited[i] = true;
          level[i] = level[s] + 1;
          queue.add(i);
        }
      }
 
      count = 0;
      for (int i = 0; i < V; i++)
        if (level[i] == l)
          count++;
    }
    return count;
  }
}
class GFG {
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Create a graph given
    // in the above diagram
    Graph g = new Graph(6);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 3);
    g.addEdge(2, 4);
    g.addEdge(2, 5);
    int level = 2;
    System.out.print(g.BFS(0, level));
  }
}
 
// This code is contributed by Rajput-Ji


Python3




# Python3 program to print
# count of nodes at given level.
from collections import deque
  
adj = [[] for i in range(1001)]
  
def addEdge(v, w):
     
    # Add w to v’s list.
    adj[v].append(w)
  
    # Add v to w's list.
    adj[w].append(v)
  
def BFS(s, l):
     
    V = 100
     
    # Mark all the vertices
    # as not visited
    visited = [False] * V
    level = [0] * V
  
    for i in range(V):
        visited[i] = False
        level[i] = 0
  
    # Create a queue for BFS
    queue = deque()
  
    # Mark the current node as
    # visited and enqueue it
    visited[s] = True
    queue.append(s)
    level[s] = 0
  
    while (len(queue) > 0):
         
        # Dequeue a vertex from
        # queue and prit
        s = queue.popleft()
        #queue.pop_front()
  
        # Get all adjacent vertices
        # of the dequeued vertex s.
        # If a adjacent has not been
        # visited, then mark it
        # visited and enqueue it
        for i in adj[s]:
            if (not visited[i]):
  
                # Setting the level
                # of each node with
                # an increment in the
                # level of parent node
                level[i] = level[s] + 1
                visited[i] = True
                queue.append(i)
  
    count = 0
    for i in range(V):
        if (level[i] == l):
            count += 1
             
    return count
  
# Driver code
if __name__ == '__main__':
     
    # Create a graph given
    # in the above diagram
    addEdge(0, 1)
    addEdge(0, 2)
    addEdge(1, 3)
    addEdge(2, 4)
    addEdge(2, 5)
  
    level = 2
  
    print(BFS(0, level))
     
# This code is contributed by mohit kumar 29


C#




// C# program to print count of nodes
// at given level.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
 
// This class represents
// a directed graph
// using adjacency
// list representation
class Graph{
 
// No. of vertices    
private int _V;
 
LinkedList<int>[] _adj;
 
public Graph(int V)
{
    _adj = new LinkedList<int>[V];
 
    for(int i = 0; i < _adj.Length; i++)
    {
        _adj[i] = new LinkedList<int>();
    }
    _V = V;
}
 
public void AddEdge(int v, int w)
{
     
    // Add w to v’s list.
    _adj[v].AddLast(w);
}
 
public int BreadthFirstSearch(int s,int l)
{
     
    // Mark all the vertices
    // as not visited
    bool[] visited = new bool[_V];
    int[] level = new int[_V];
     
    for(int i = 0; i < _V; i++)
    {
        visited[i] = false;
        level[i] = 0;
    }
     
    // Create a queue for BFS
    LinkedList<int> queue = new LinkedList<int>();
     
    // Mark the current node as
    // visited and enqueue it
    visited[s] = true;
    level[s] = 0;
    queue.AddLast(s);        
 
    while(queue.Any())
    {
         
        // Dequeue a vertex from
        // queue and print it
        s = queue.First();
         
        // Console.Write( s + " " );
        queue.RemoveFirst();
 
        LinkedList<int> list = _adj[s];
 
        foreach(var val in list)            
        {
            if (!visited[val])
            {
                visited[val] = true;
                level[val] = level[s] + 1;
                queue.AddLast(val);
            }
        }
    }
     
    int count = 0;
    for(int i = 0; i < _V; i++)
        if (level[i] == l)
            count++;
             
    return count;
}
}
 
// Driver code
class GFG{
     
static void Main(string[] args)
{
     
    // Create a graph given
    // in the above diagram
    Graph g = new Graph(6);
 
    g.AddEdge(0, 1);
    g.AddEdge(0, 2);
    g.AddEdge(1, 3);
    g.AddEdge(2, 4);
    g.AddEdge(2, 5);
 
    int level = 2;
     
    Console.WriteLine(g.BreadthFirstSearch(0, level));
}
}
 
// This code is contributed by anvudemy1


Output: 
 

 3

 

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