Description
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
nelements is the sum of thenelements divided bynand rounded down to the nearest integer. - A subtree of
rootis a tree consisting ofrootand 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
Solutions
Code/performance improvements.
Language: javascript(2026-09-10 09:34)DONE
CPU Performance99.20%
Memory Performance60.16%
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var averageOfSubtree = function(root) {
let res = 0;
const dfs = (node) => {
if (!node) return { sum: 0, count: 0 };
const left = dfs(node.left);
const right = dfs(node.right);
const sum = node.val + left.sum + right.sum;
const count = 1 + left.count + right.count;
if (Math.floor(sum / count) === node.val) {
res++;
}
return { sum, count };
};
dfs(root);
return res;
};