Skip to main content
Talentd

Count Nodes Equal to Average of Subtree - Solution & Explanation

MediumTreeDepth-First SearchBinary Tree14 min readAsked at: Amazon, Meta, Snowflake +1
Practice this problem

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.
  • 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

Approach Overview

Problem Overview: You are given the root of a binary tree. For every node, compute the average value of all nodes in its subtree (including itself). Count how many nodes have a value equal to the integer average of their subtree.

Approach 1: Recursion with Post-Order Traversal (O(n) time, O(h) space)

This approach uses a post-order Depth-First Search to process children before their parent. Each recursive call returns two values: the sum of the subtree and the number of nodes in that subtree. With these values, you compute average = sum // count and compare it with the current node's value. Because each node is visited exactly once and aggregation happens while returning from recursion, the algorithm runs in O(n) time with O(h) space for the recursion stack, where h is the tree height. This pattern is common in tree problems where subtree metrics must be combined bottom-up.

Approach 2: Iterative DFS Using Stack (O(n) time, O(n) space)

An iterative solution simulates the same post-order traversal using an explicit stack instead of recursion. Each stack frame stores the node and a state flag indicating whether its children have already been processed. A hash map (or dictionary) stores computed (sum, count) pairs for each node so the parent can aggregate them later. Once both children are processed, compute the subtree sum and node count, then check whether the node value equals sum // count. This method avoids recursion depth limits but uses O(n) auxiliary space for the stack and cached results.

Recommended for interviews: The recursive post-order traversal is the expected solution. Interviewers want to see that you recognize subtree computations require bottom-up aggregation, a common binary tree pattern. The iterative stack version demonstrates deeper understanding of DFS mechanics but is usually unnecessary unless recursion limits are a concern.

Approach 1: Recursion with Post-Order Traversal

This approach involves using a post-order traversal (left, right, root) to compute the sum and count of nodes for each subtree. At each node, the sum of values and the total count of the subtree nodes are determined. The average is calculated by dividing the sum by the count and compared with the node's value. If they are equal, a counter is incremented.

This Python solution defines a recursive DFS function `dfs` that traverses the tree in a post-order manner. For each node, it computes the sum and count of the subtree rooted at that node. If the node value equals the average (sum // count) of its subtree, the counter `self.count` is incremented.

Code

PythonJavaScript

Python

JavaScript

Complexity

Time Complexity: O(n), where n is the number of nodes in the tree, as each node is visited once.
Space Complexity: O(h), where h is the height of the tree, which accounts for the recursion stack.

Try this approach in the editor →

Approach 2: Iterative Using Stack

This approach involves using an iterative method with an explicit stack to perform a post-order traversal. Similar to the recursive method, it calculates the sum and count of the subtree nodes for each node using an additional data structure to simulate recursion.

The C# solution utilizes an explicit stack to perform a post-order traversal. It uses a dictionary to store the sum and count for each node. The node is processed twice: first to traverse its children and then to calculate its subtree's sum and count.

Code

C#Java

C#

Java

Complexity

Time Complexity: O(n), since each node is processed twice.
Space Complexity: O(n), for the stack and the dictionary storing sums and counts for each node.

Try this approach in the editor →

Approach 3: DFS

We design a function dfs, which calculates the sum and the number of nodes of the subtree rooted at the current node.

The execution process of the function dfs is as follows:

  • If the current node is null, return (0, 0).
  • Otherwise, we recursively calculate the sum and the number of nodes of the left and right subtrees, denoted as (ls, ln) and (rs, rn), respectively. Then, the sum s and the number of nodes n of the subtree rooted at the current node are ls + rs + root.val and ln + rn + 1, respectively. If s / n = root.val, it means the current node meets the requirement of the problem, and we increment the answer ans by 1.
  • Finally, the function dfs returns s and n.

We initialize the answer ans to 0, then call the dfs function, and finally return the answer ans.

The time complexity is O(n), and the space complexity is O(n). Here, n represents the number of nodes in the binary tree.

Code

PythonJavaC++GoTypeScript

Python

Java

C++

Go

TypeScript

Try this approach in the editor →

Complexity Comparison

ApproachComplexity
Recursion with Post-Order Traversal

Time Complexity: O(n), where n is the number of nodes in the tree, as each node is visited once.
Space Complexity: O(h), where h is the height of the tree, which accounts for the recursion stack.

Iterative Using Stack

Time Complexity: O(n), since each node is processed twice.
Space Complexity: O(n), for the stack and the dictionary storing sums and counts for each node.

DFS—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Recursion with Post-Order TraversalO(n)O(h)Best general solution. Clean bottom-up aggregation of subtree sum and count.
Iterative DFS Using StackO(n)O(n)When recursion depth may overflow or when implementing DFS iteratively.

Video Solution

Count Nodes Equal to Average of Subtree | 2 Approaches | Leetcode-2265 | Explanation ➕ Live Coding • codestorywithMIK • 7,344 views views

Watch 9 more video solutions →

Frequently Asked Questions

Is Count Nodes Equal to Average of Subtree easy or hard?
The problem is rated Medium because it requires recognizing a post-order traversal pattern and returning multiple values from each DFS call. The implementation is straightforward once you realize subtree sum and count must be computed before evaluating the parent node.
Count Nodes Equal to Average of Subtree Python/Java solution
Most implementations use a recursive helper that returns (sum, count) for each subtree. Python and JavaScript solutions typically implement concise DFS recursion, while Java and C# may also use iterative stack-based DFS to simulate post-order traversal.
How to solve Count Nodes Equal to Average of Subtree in O(n)?
Use post-order DFS to compute subtree statistics bottom-up. Each recursive call returns the subtree sum and node count. After processing left and right children, calculate the average using sum // count and compare it with the node value. Since each node contributes to exactly one calculation, the algorithm runs in O(n).
What is the best approach for Count Nodes Equal to Average of Subtree?
The optimal approach is post-order DFS traversal. For each node, compute the sum and number of nodes in its subtree, then check if node.val equals sum // count. Each node is processed once, giving O(n) time complexity with O(h) recursion space where h is the tree height.
Is Count Nodes Equal to Average of Subtree asked at Google/Amazon/Meta?
Tree aggregation problems using DFS appear frequently in interviews at companies like Amazon, Google, and Meta. Variants that require computing subtree sums, averages, or counts are common because they test understanding of post-order traversal and recursion patterns.
What data structure is used in Count Nodes Equal to Average of Subtree?
The main data structure is a binary tree. The algorithm relies on depth-first search traversal. The recursive version uses the call stack, while the iterative approach uses an explicit stack and optionally a hash map to store subtree results.
What is the time complexity of Count Nodes Equal to Average of Subtree?
The optimal solution runs in O(n) time because every node in the binary tree is visited exactly once during the DFS traversal. Space complexity is O(h) for recursion stack depth, where h is the height of the tree (O(log n) for balanced trees and O(n) for skewed trees).

Ready to solve this problem?

Practice Count Nodes Equal to Average of Subtree with our built-in code editor and test cases.

Practice on FleetCode
Previous

Count Nodes Equal to Sum of Descendants

Next

Insufficient Nodes in Root to Leaf Paths