What is the time complexity of Count Nodes Equal to Average of Subtree?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Nodes Equal to Average of Subtree?

This problem uses Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Nodes Equal to Average of Subtree?

Insight 1: A post-order traversal is ideal because we need to know the sum and count of the left and right subtrees before processing the current node. Insight 2: The core logic involves recursively calculating the sum and count of each subtree and then checking if the current node's value matches the calculated average.

Which companies ask LeetCode 2265. Count Nodes Equal to Average of Subtree in interviews?

This problem is frequently asked at INDmoney, Snowflake.

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Count Nodes Equal to Average of Subtree - LeetCode 2265 Solution

Count Nodes Equal to Average of Subtree - Complete Solution Guide

Count Nodes Equal to Average of Subtree is LeetCode problem 2265, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. F

Detailed Explanation

The problem asks us to traverse a binary tree and, for each node, determine if the node's value is equal to the average of all the values in its subtree (including the node itself). The average is calculated by summing all the values in the subtree, dividing by the number of nodes in the subtree, and rounding down to the nearest integer. The input is the root of the binary tree, and the output is the number of nodes that satisfy this condition.

Solution Approach

The provided solutions use a post-order Depth-First Search (DFS) traversal. The `postOrder` function (or equivalent in other languages) recursively calculates the sum and the number of nodes in each subtree. For each node, it adds its own value to the sum of its left and right subtrees and increments the node count by the number of nodes in its left and right subtrees, plus one (for the node itself). It then calculates the average of the subtree and checks if the average equals the node's value. If it does, it increments a counter.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0. This counter will store the number of nodes satisfying the average condition.
Step 2: Define a recursive function `postOrder(node)` that takes a node as input and returns a tuple (or array) containing the sum of node values and the number of nodes in the subtree rooted at `node`.
Step 3: Base case for recursion: If `node` is null, return (0, 0) as there are no nodes and the sum is 0.
Step 4: Recursively call `postOrder(node.left)` to get the sum and count of the left subtree.
Step 5: Recursively call `postOrder(node.right)` to get the sum and count of the right subtree.
Step 6: Calculate the `currentSum` by adding the node's value, the sum of the left subtree, and the sum of the right subtree.
Step 7: Calculate the `currentCount` by adding 1 (for the current node), the count of the left subtree, and the count of the right subtree.
Step 8: Calculate the `average` by dividing `currentSum` by `currentCount` (integer division).
Step 9: Check if `average` equals the node's value. If it does, increment the `count`.
Step 10: Return `currentSum` and `currentCount` as a tuple (or array).
Step 11: Call `postOrder(root)` to start the traversal from the root of the tree.
Step 12: Return the final `count`.

Key Insights

Insight 1: A post-order traversal is ideal because we need to know the sum and count of the left and right subtrees before processing the current node.
Insight 2: The core logic involves recursively calculating the sum and count of each subtree and then checking if the current node's value matches the calculated average.
Insight 3: We can use a simple integer division to get the rounded-down average, as required by the problem statement.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

This problem involves: Tree, Depth-First Search, Binary Tree.

Companies

Asked at: INDmoney, Snowflake.

Count Nodes Equal to Average of Subtree | LeetCode 2265

Medium

Tree Depth-First Search Binary Tree

Time: O(n)Space: O(h)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int count = 0;

    public int averageOfSubtree(TreeNode root) {
        count = 0;
        postOrder(root);
        return count;
    }

    private int[] postOrder(TreeNode node) {
        if (node == null) {
            return new int[]{0, 0}; // sum, number of nodes
        }

        int[] leftResult = postOrder(node.left);
        int[] rightResult = postOrder(node.right);

        int leftSum = leftResult[0];
        int leftCount = leftResult[1];
        int rightSum = rightResult[0];
        int rightCount = rightResult[1];

        int currentSum = node.val + leftSum + rightSum;
        int currentCount = 1 + leftCount + rightCount;

        int average = currentSum / currentCount;

        if (average == node.val) {
            count++;
        }

        return new int[]{currentSum, currentCount};
    }
}

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Problem Info

Description

Problem 2265: Count Nodes Equal to Average of Subtree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

INDmoneySnowflake

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2265

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Initialize a counter `count` to 0. This counter will store the number of nodes satisfying the average condition.
2
Step 2: Define a recursive function `postOrder(node)` that takes a node as input and returns a tuple (or array) containing the sum of node values and the number of nodes in the subtree rooted at `node`.
3
Step 3: Base case for recursion: If `node` is null, return (0, 0) as there are no nodes and the sum is 0.
4
Step 4: Recursively call `postOrder(node.left)` to get the sum and count of the left subtree.
5
Step 5: Recursively call `postOrder(node.right)` to get the sum and count of the right subtree.
6
Step 6: Calculate the `currentSum` by adding the node's value, the sum of the left subtree, and the sum of the right subtree.
7
Step 7: Calculate the `currentCount` by adding 1 (for the current node), the count of the left subtree, and the count of the right subtree.
8
Step 8: Calculate the `average` by dividing `currentSum` by `currentCount` (integer division).
9
Step 9: Check if `average` equals the node's value. If it does, increment the `count`.
10
Step 10: Return `currentSum` and `currentCount` as a tuple (or array).
11
Step 11: Call `postOrder(root)` to start the traversal from the root of the tree.
12
Step 12: Return the final `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(h)

📝

Why?

The time complexity is O(n) because we perform a post-order traversal, which requires visiting each node once. The space complexity is O(h) because the maximum depth of the recursion stack is equal to the height of the tree.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative DFS

A similar approach can be implemented using an iterative DFS with a stack. This approach avoids recursion and might reduce the space complexity in some cases.

Time: O(n)Space: O(n) in the worst case (skewed tree), as the stack might contain all nodes.

Trade-offs: Iterative DFS can be slightly more complex to implement but may provide better control over memory usage in certain environments or for very deep trees. It avoids potential stack overflow issues that can arise in extremely unbalanced trees during recursion.

✨

Key Insights

1Insight 1: A post-order traversal is ideal because we need to know the sum and count of the left and right subtrees before processing the current node.
2Insight 2: The core logic involves recursively calculating the sum and count of each subtree and then checking if the current node's value matches the calculated average.
3Insight 3: We can use a simple integer division to get the rounded-down average, as required by the problem statement.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: A tree with only one node. The algorithm correctly handles this because the sum and count are simply the node's value and 1, respectively, and the average will be equal to the node's value.
Edge case 2: Empty tree (null root). While the problem states the number of nodes is in [1, 1000], handling a null root gracefully is good practice. In this case, the main function's return would not execute the traversal and return initial 0 (for no nodes matching the average condition).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the average. Forgetting to include the current node's value in the sum or incorrectly calculating the number of nodes in the subtree.

✗Common mistake 2: Not handling the base case (null node) properly, leading to errors in the recursive calls.

✗Common mistake 3: Using pre-order or in-order traversal instead of post-order. These traversals do not readily provide the subtree sum and count before processing the current node.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing network traffic patterns. Each node could represent a router, and the value could represent the traffic passing through it. The algorithm can identify routers where the local traffic matches the average traffic in its subtree, potentially indicating congestion or unusual activity.

🏦Real-world scenario 2: Analyzing organizational hierarchies. Each node could represent an employee, and the value could represent their performance score. The algorithm can identify employees whose performance matches the average performance of their team and sub-teams, which can be useful for performance reviews or identifying high-potential employees.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Nodes Equal to Average of Subtree"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?