Count Nodes Equal to Average of Subtree — LeetCode 2265 Python Solution

The 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.

Example

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.

Python solution

class Solution:
    def averageOfSubtree(self, root: TreeNode) -> int:
        def dfs(root) -> tuple:
            if not root:
                return 0, 0
            ls, ln = dfs(root.left)
            rs, rn = dfs(root.right)
            s = ls + rs + root.val
            n = ln + rn + 1
            nonlocal ans
            ans += int(s // n == root.val)
            return s, n

        ans = 0
        dfs(root)
        return ans

Complexity

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 2265. Count Nodes Equal to Average of Subtree is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

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