What is the time complexity of Count Complete Tree Nodes?

The optimal solution has a time complexity of O((log(n))^2).

What is the space complexity of Count Complete Tree Nodes?

The space complexity is O(log(n)).

What data structures are used to solve Count Complete Tree Nodes?

This problem uses Binary Search, Bit Manipulation, Tree, Binary Tree. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Complete Tree Nodes?

Insight 1: A complete binary tree's left and right subtrees might have different heights. This difference is crucial in determining the algorithm's strategy. Insight 2: If the left and right subtrees have the same height, the tree is a perfect binary tree (or full up to the second to last level), allowing for a quick calculation of the total nodes using bit manipulation (2^(height+1) -1).

Which companies ask LeetCode 222. Count Complete Tree Nodes in interviews?

This problem is frequently asked at Dunzo.

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Count Complete Tree Nodes - LeetCode 222 Solution

Count Complete Tree Nodes - Complete Solution Guide

Count Complete Tree Nodes is LeetCode problem 222, a Easy 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 complete binary tree, return the number of the nodes in the tree. According to Wikipedia , every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2 h nodes inclusive at the last level h . Design an algorithm that runs in less than O(n) time complexity. Example 1: Input: root = [1,2,3,4,5,6] Output: 6 Example 2: Input: root = [] Output: 0 Example 3: Input: roo

Detailed Explanation

The problem asks you to count the number of nodes in a complete binary tree. A complete binary tree is a binary tree in which all levels are completely filled except possibly the last level, and the last level has all keys as left as possible. The input is the root of the complete binary tree, and the output is the total number of nodes in the tree. The challenge is to design an algorithm that runs in less than O(n) time complexity, where n is the number of nodes.

Solution Approach

The provided solution employs a recursive divide-and-conquer strategy. It first checks if the left and right subtrees have the same height. If they do, it means the left subtree is a perfect binary tree. In this case, we can efficiently calculate the total nodes without traversing the entire subtree. If the heights are different, it means the right subtree is not a perfect binary tree, so it recursively counts nodes in both left and right subtrees and adds 1 (for the root node).

Step-by-Step Algorithm

Step 1: Check if the root is null. If it is, return 0 (empty tree).
Step 2: Calculate the height of the left subtree and the height of the right subtree by traversing down to the leftmost and rightmost leaves respectively.
Step 3: If the left and right subtree heights are equal, the tree is essentially a perfect binary tree (or a perfect binary tree up to the second to last level). Calculate the total number of nodes using the formula (1 << (leftHeight + 1)) - 1 (or (1 << (rightHeight + 1)) - 1; both are equivalent in this case).
Step 4: If the left and right subtree heights are unequal, the total number of nodes is 1 (the root) plus the number of nodes in the left subtree plus the number of nodes in the right subtree. Recursively call the `countNodes` function on the left and right subtrees to count their nodes.
Step 5: Return the total count of nodes.

Key Insights

Insight 1: A complete binary tree's left and right subtrees might have different heights. This difference is crucial in determining the algorithm's strategy.
Insight 2: If the left and right subtrees have the same height, the tree is a perfect binary tree (or full up to the second to last level), allowing for a quick calculation of the total nodes using bit manipulation (2^(height+1) -1).
Insight 3: A recursive approach, combined with the perfect binary tree check, optimizes the traversal by avoiding unnecessary recursive calls on the perfectly balanced portion of the tree.

Complexity Analysis

Time Complexity: O((log(n))^2)

Space Complexity: O(log(n))

Topics

This problem involves: Binary Search, Bit Manipulation, Tree, Binary Tree.

Companies

Asked at: Dunzo.

Count Complete Tree Nodes | LeetCode 222

Easy

Binary Search Bit Manipulation Tree Binary Tree

Time: O((log(n))^2)Space: O(log(n))

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 {
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        int leftHeight = 0, rightHeight = 0;
        TreeNode left = root, right = root;
        while (left != null) {
            leftHeight++;
            left = left.left;
        }
        while (right != null) {
            rightHeight++;
            right = right.right;
        }
        if (leftHeight == rightHeight) return (1 << leftHeight) - 1;
        return 1 + countNodes(root.left) + countNodes(root.right);
    }
}

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Problem Info

Description

Problem 222: Count Complete Tree Nodes - Detailed explanation and solution approaches.

Topics

Binary Search Bit Manipulation Tree Binary Tree

Companies

Dunzo

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:222

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

5 steps

1
Step 1: Check if the root is null. If it is, return 0 (empty tree).
2
Step 2: Calculate the height of the left subtree and the height of the right subtree by traversing down to the leftmost and rightmost leaves respectively.
3
Step 3: If the left and right subtree heights are equal, the tree is essentially a perfect binary tree (or a perfect binary tree up to the second to last level). Calculate the total number of nodes using the formula (1 << (leftHeight + 1)) - 1 (or (1 << (rightHeight + 1)) - 1; both are equivalent in this case).
4
Step 4: If the left and right subtree heights are unequal, the total number of nodes is 1 (the root) plus the number of nodes in the left subtree plus the number of nodes in the right subtree. Recursively call the `countNodes` function on the left and right subtrees to count their nodes.
5
Step 5: Return the total count of nodes.

💡

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((log(n))^2)

💾

Space Complexity

O(log(n))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is not strictly O(n) because of the recursive calls. However, it is significantly better than O(n) for complete binary trees. In the worst case, the recursive calls only descend through the unbalanced portion of the tree which has a logarithmic height (log n). Finding the height of each subtree takes O(log n) time, and this height calculation is performed at most log n times. Space complexity comes from the recursive call stack, which can go up to the height of the tree (log n).

📖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

Level Order Traversal

Traverse the tree level by level using a queue. Count the nodes as you visit them. This is a straightforward approach but has a time complexity of O(n).

Time: O(n)Space: O(w), where w is the maximum width of the tree. In a complete binary tree, this is O(n) in the worst case.

Trade-offs: Simpler to implement but less efficient than the provided solution for large complete binary trees.

✨

Key Insights

1Insight 1: A complete binary tree's left and right subtrees might have different heights. This difference is crucial in determining the algorithm's strategy.
2Insight 2: If the left and right subtrees have the same height, the tree is a perfect binary tree (or full up to the second to last level), allowing for a quick calculation of the total nodes using bit manipulation (2^(height+1) -1).
3Insight 3: A recursive approach, combined with the perfect binary tree check, optimizes the traversal by avoiding unnecessary recursive calls on the perfectly balanced portion of the tree.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root == null). The function correctly returns 0.
Edge case 2: Tree with only one node. The function correctly returns 1.
Edge case 3: A perfectly balanced binary tree. The algorithm efficiently calculates the node count using the bit manipulation formula, avoiding unnecessary recursion.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that all subtrees are perfectly balanced and directly applying the (2^(h+1)-1) formula without checking for height equality. This leads to incorrect results for unbalanced complete binary trees.

✗Common mistake 2: Not handling the base case (empty tree) properly, resulting in null pointer exceptions or incorrect results.

✅

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 the structure and size of hierarchical data, such as file systems or organizational charts represented as complete binary trees. This algorithm can efficiently determine the total number of files or employees.

🏦Real-world scenario 2: In game development or computer graphics, complete binary trees can represent spatial data structures (like quadtrees or octrees). This algorithm can quickly count the number of objects or regions in the space.

💡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

Binary search isn't just for sorted arrays. It can be applied to any monotonic function. Look for the "search space" in the problem.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Integer overflow in (left + right) / 2. Use left + (right - left) / 2 instead. Also watch for infinite loops.

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: July 13, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Complete Tree Nodes"?

For this easy-level Binary Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?