Finding the number of nodes at each level during bfs
<p>You have been given a Tree consisting of N nodes. A tree is a fully-connected graph consisting of N nodes and N−1 edges. The nodes in this tree are indexed from 1 to N. Consider node indexed 1 to be the root node of this tree. The root node lies at level one in the tree. You shall be given the tree and a single integer x . You need to find out the number of nodes lying on level x.</p>

<p>Input Format</p>

<p>The first line consists of a single integer N denoting the number of nodes in the tree. Each of the next n−1 lines consists of 2 integers a andb denoting an undirected edge between node a and node b. The next line consists of a single integer x.</p>

<p>Output Format</p>

<p>You need to print a single integers denoting the number of nodes on level x.</p>

<hr>

<p>below is the code snippet i wrote and its not correct. Please help me find the error.</p>

<pre><code>import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Scanner;

public class LevelNodes {
    public static ArrayList[] adj;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        boolean[] visites = new boolean[N];
        ArrayList[] adj = new ArrayList[N];
        for(int j=0;j&lt;N;j++){
            adj[j] = new ArrayList();
        }
        for(int i = 0;i &lt; (N-1) ;i++){
            int a = sc.nextInt();
            int b = sc.nextInt();
            adj[a-1].add(b);
            adj[b-1].add(a);
        }
        int level = sc.nextInt();
        int counter = 0;
        LinkedList list = new LinkedList();
        list.add(1);
        counter++;
        while(!list.isEmpty()){
            int n = (Integer) list.poll();
            for(int x = 0; x&lt; adj[n-1].size();x++){
                list.add(adj[n-1].get(x));
            }
            counter++;
            if(counter == level){
                System.out.println(list.size());
                break;
            }

        }
    }

}
</code></pre>