2265. Count Nodes Equal to Average of Subtree

T: O(n)
S: O(n)

```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;
 *     }
 * }

 leaf node is ok

 post order
 */
class Solution {
    private int result = 0;
    public int averageOfSubtree(TreeNode root) {
        dfs(root);
        return result;
    }
    private int[] dfs(TreeNode node) {
        if (node == null) {
            return new int[] {0, 0};
        }
        int[] left = dfs(node.left);
        int[] right = dfs(node.right);

        int currentSum = left[0] + right[0] + node.val;
        int currentCount = left[1] + right[1] + 1;
        if ((currentSum/currentCount) == node.val) {
            result++;
        }
        return new int[] {currentSum, currentCount};
    }
}

/***
T: O(n)
S: O(n)

 */
```