Leetcode #2265: Count Nodes Equal to Average of Subtree
In this guide, we solve Leetcode #2265 Count Nodes Equal to Average of Subtree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
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.
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.
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
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.