How do you solve Count Nodes Equal to Average of Subtree on LeetCode?

Count Nodes Equal to Average of Subtree (LeetCode #2265) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Understanding how to calculate the average requires both the sum and the count of nodes in a subtree.

What is the brute force solution for Count Nodes Equal to Average of Subtree?

The brute force approach for Count Nodes Equal to Average of Subtree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the sum and count of nodes for every node in the tree. This means for each node, we traverse its entire subtree, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Nodes Equal to Average of Subtree?

Count Nodes Equal to Average of Subtree uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Count Nodes Equal to Average of Subtree?

Always clarify how you will calculate averages and ensure you understand integer division. Discuss the time complexity of your approach and be prepared to optimize if necessary. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Nodes Equal to Average of Subtree?

Not considering integer division when calculating the average, which can lead to incorrect comparisons. Forgetting to return both sum and count from the DFS function.

#2265

Count Nodes Equal to Average of Subtree

Medium

TreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal approach uses a single depth-first search (DFS) to calculate both the sum and count of nodes in each subtree in one traversal, making it much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS on the tree, calculating the sum and count of nodes for each subtree.
2Step 2: For each node, check if its value equals the average of its subtree using the sum and count obtained.
3Step 3: Count the nodes that satisfy the condition.

solution.py23 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def averageOfSubtree(self, root: TreeNode) -> int:
10        def dfs(node):
11            if not node:
12                return (0, 0)
13            left_sum, left_count = dfs(node.left)
14            right_sum, right_count = dfs(node.right)
15            total_sum = left_sum + right_sum + node.val
16            total_count = left_count + right_count + 1
17            if node.val == total_sum // total_count:
18                self.count += 1
19            return (total_sum, total_count)
20
21        self.count = 0
22        dfs(root)
23        return self.count

ℹ

Complexity note: The time complexity is linear because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Understanding how to calculate the average requires both the sum and the count of nodes in a subtree.
2Using a single DFS traversal can significantly improve efficiency compared to recalculating sums for each node.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.