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2265. Count Nodes Equal to Average of Subtree

Difficulty Topics

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 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

count-nodes-equal-to-average-of-subtree.py
# 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:
        res = 0

        def go(node):
            if not node: return (0, 0)

            lCount, lValues = go(node.left)
            rCount, rValues = go(node.right)

            nodeCount, nodeValues = 1 + lCount + rCount, node.val + lValues + rValues

            average = nodeValues // nodeCount

            if node.val == average:
                nonlocal res

                res += 1

            return (nodeCount, nodeValues)

        go(root)

        return res