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

/**
 * 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) {
    return helper(root).res;
};

var helper = function(root) {
    if (!root) return { sum: 0, num: 0, res: 0 };
    var left = helper(root.left);
    var right = helper(root.right);
    var sum = left.sum + right.sum + root.val;
    var num = left.num + right.num + 1;
    var res = left.res + right.res + (Math.floor(sum / num) === root.val ? 1 : 0);
    return { sum, num, res };
};

Explain:

nope.

Complexity:

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