README.md

2265. Count Nodes Equal to Average of Subtree

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

Solutions (Python)

1. Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def averageOfSubtree(self, root: Optional[TreeNode]) -> int:
        return self.dfs(root)[2]

    def dfs(self, root: TreeNode) -> (int, int, int):
        if not root:
            return (0, 0, 0)

        count_left, sum_left, ret_left = self.dfs(root.left)
        count_right, sum_right, ret_right = self.dfs(root.right)
        count_root = count_left + count_right + 1
        sum_root = sum_left + sum_right + root.val
        ret_root = ret_left + ret_right + \
            (1 if root.val == sum_root // count_root else 0)

        return (count_root, sum_root, ret_root)