LeetCode-in-Java

2265. Count Nodes Equal to Average of Subtree

Medium

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.

Note:

Example 1:

Input: root = [4,8,5,0,1,null,6]

Output: 5

Explanation:

For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.

For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.

For the node with value 0: The average of its subtree is 0 / 1 = 0.

For the node with value 1: The average of its subtree is 1 / 1 = 1.

For the node with value 6: The average of its subtree is 6 / 1 = 6.

Example 2:

Input: root = [1]

Output: 1

Explanation: For the node with value 1: The average of its subtree is 1 / 1 = 1.

Constraints:

Solution

import com_github_leetcode.TreeNode;

public class Solution {
    private int ans = 0;

    public int averageOfSubtree(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int[] dfs(TreeNode node) {
        if (node == null) {
            return new int[] {0, 0};
        }
        int[] left = dfs(node.left);
        int[] right = dfs(node.right);
        int currsum = left[0] + right[0] + node.val;
        int currcount = left[1] + right[1] + 1;
        if (currsum / currcount == node.val) {
            ++ans;
        }
        return new int[] {currsum, currcount};
    }
}