Obtain forest out of tree with even number of nodes

I'm stuck on a code challenge, and I want a hint .

PROBLEM : You are given a tree data structure (without cycles) and are asked to remove as many "edges" (connections) as possible, creating smaller trees with even numbers of nodes. This problem is always solvable as there are an even number of nodes and connections.

Your task is to count the removed edges.

Input: The first line of input contains two integers N and M. N is the number of vertices and M is the number of edges. 2 <= N <= 100. Next M lines contains two integers ui and vi which specifies an edge of the tree. (1-based index)

Output: Print the number of edges removed.

Sample Input

10 9
2 1
3 1
4 3
5 2
6 1
7 2
8 6
9 8
10 8

Sample Output : 2

Explanation : On removing the edges (1, 3) and (1, 6), we can get the desired result.

---

I used BFS to travel through the nodes. First, maintain an array separately to store the total number of child nodes + 1. So, you can initially assign all the leaf nodes with value 1 in this array. Now start from the last node and count the number of children for each node. This will work in bottom to top manner and the array that stores the number of child nodes will help in runtime to optimize the code.

Once you get the array after getting the number of children nodes for all the nodes, just counting the nodes with even number of nodes gives the answer. Note: I did not include root node in counting in final step.

---

This is my solution. I didn't use bfs tree, just allocated another array for holding eachnode's and their children nodes total number.

import java.util.Scanner;
import java.util.Arrays;

public class Solution {
        public static void main(String[] args) {
                int tree[];
                int count[];

                Scanner scan = new Scanner(System.in);

                int N = scan.nextInt(); //points
                int M = scan.nextInt();

                tree = new int[N];
                count = new int[N];
                Arrays.fill(count, 1);

                for(int i=0;i<M;i++)
                {
                        int u1 = scan.nextInt();
                    int v1 = scan.nextInt();

                    tree[u1-1] = v1;

                    count[v1-1] += count[u1-1];

                    int root = tree[v1-1];

                    while(root!=0)
                    {
                        count[root-1] += count[u1-1];
                        root = tree[root-1];
                    }
                }

                System.out.println("");

                int counter = -1;
                for(int i=0;i<count.length;i++)
                {
                        if(count[i]%2==0)
                        {
                                counter++;
                        }

                }
                System.out.println(counter);

        }

}

---

If you observe the input, you can see that it is quite easy to count the number of nodes under each node. Consider (ab) as the edge input, in every case, a is the child and b is the immediate parent. The input always has edges represented bottom-up.

So its essentially the number of nodes which have an even count(Excluding the root node). I submitted the below code on Hackerrank and all the tests passed. I guess all the cases in the input satisfy the rule.

def find_edges(count):
    root = max(count)

    count_even = 0

    for cnt in count:
        if cnt % 2 == 0:
            count_even += 1

    if root % 2 == 0:
        count_even -= 1

    return count_even

def count_nodes(edge_list, n, m):
    count = [1 for i in range(0, n)]

    for i in range(m-1,-1,-1):
        count[edge_list[i][1]-1] += count[edge_list[i][0]-1]

return find_edges(count)

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