2265. Count Nodes Equal to Average of Subtree

Problem

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:

• The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer.

• A subtree of root is a tree consisting of root and all of its descendants.

  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:

• The number of nodes in the tree is in the range [1, 1000].

• 0 <= Node.val <= 1000

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
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};
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).