What is the time complexity of Balanced Binary Tree?

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

What data structures are used to solve Balanced Binary Tree?

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

What are the key insights for solving Balanced Binary Tree?

**Bottom-Up Balance Validation with Height Propagation:** The problem's recursive nature—every node's subtrees must be balanced—is perfectly addressed by a bottom-up (post-order) DFS. Each node is responsible for not only computing its own subtree's height but also verifying its own balance condition *after* its children have reported their status. This ensures that an imbalance deep in the tree is detected and handled before higher-level nodes attempt their own checks. **Sentinel Value for Early Exit (Short-Circuiting):** The use of `-1` as a sentinel value is the critical optimization. Instead of a function that *only* returns height and then needing a separate boolean flag or a second pass to check balance, the `height` function itself communicates both states. If any recursive call encounters an imbalance or receives a `-1` from its children, it immediately propagates `-1` upwards. This effectively "short-circuits" the computation for that entire branch, preventing unnecessary calculations once an imbalance is confirmed.

Which companies ask LeetCode 110. Balanced Binary Tree in interviews?

This problem is frequently asked at Adobe, Apple, Capgemini, Meta, Uber and 1 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Balanced Binary Tree - LeetCode 110 Solution

Q: How to solve Balanced Binary Tree?

The provided solution employs a clever depth-first search (DFS) approach that combines two critical tasks into a single recursive function: calculating the height of a subtree and simultaneously verifying its balance property. The `height` function traverses the tree in a post-order manner. For any given `node`, it first recursively determines the height and balance status of its left and right children. The elegance here is how an imbalance is signaled: if at any point a subtree (either left or right) is found to be unbalanced, its `height` call returns a special sentinel value, `-1`. This `-1` isn't a valid height, so when a parent node receives `-1` from either of its children, it immediately knows that an imbalance exists further down its branch. It then propagates this `-1` upwards, effectively short-circuiting further balance checks for that branch. If both children return valid heights, the current node then checks its own balance: `abs(left_height - right_height) > 1`. If it's imbalanced, it also returns `-1`. Otherwise, it calculates and returns its actual height `max(left_height, right_height) + 1`. The initial call `height(root)` then simply checks if the returned value is `-1`. If it is, the tree is unbalanced; otherwise, it's balanced. This strategy avoids redundant passes and ensures an efficient O(N) solution.

Balanced Binary Tree - Complete Solution Guide

Balanced Binary Tree is LeetCode problem 110, 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 a binary tree, determine if it is height-balanced . Example 1: Input: root = [3,9,20,null,null,15,7] Output: true Example 2: Input: root = [1,2,2,3,3,null,null,4,4] Output: false Example 3: Input: root = [] Output: true Constraints: The number of nodes in the tree is in the range [0, 5000] . -10 4 <= Node.val <= 10 4

Detailed Explanation

We're tasked with determining if a given binary tree is "height-balanced." At its core, this property dictates that for *every single node* within the tree, the absolute difference between the height of its left subtree and the height of its right subtree must not exceed 1. It's a recursively defined property; if even one node fails this criterion, the entire tree is considered unbalanced. This isn't merely about the root node's children; the balance condition must hold true all the way down to the leaf nodes, checking every potential subtree. This pervasive requirement is what often makes the problem tricky if one only considers the root.

Solution Approach

The provided solution employs a clever depth-first search (DFS) approach that combines two critical tasks into a single recursive function: calculating the height of a subtree and simultaneously verifying its balance property. The `height` function traverses the tree in a post-order manner. For any given `node`, it first recursively determines the height and balance status of its left and right children. The elegance here is how an imbalance is signaled: if at any point a subtree (either left or right) is found to be unbalanced, its `height` call returns a special sentinel value, `-1`. This `-1` isn't a valid height, so when a parent node receives `-1` from either of its children, it immediately knows that an imbalance exists further down its branch. It then propagates this `-1` upwards, effectively short-circuiting further balance checks for that branch. If both children return valid heights, the current node then checks its own balance: `abs(left_height - right_height) > 1`. If it's imbalanced, it also returns `-1`. Otherwise, it calculates and returns its actual height `max(left_height, right_height) + 1`. The initial call `height(root)` then simply checks if the returned value is `-1`. If it is, the tree is unbalanced; otherwise, it's balanced. This strategy avoids redundant passes and ensures an efficient O(N) solution.

Step-by-Step Algorithm

Step 1: Define a recursive helper function (e.g., `height` or `checkBalance`) that takes a tree node as input.
Step 2: Base case: If the node is null, return 0 (height of an empty subtree is 0).
Step 3: Recursively calculate the height of the left and right subtrees.
Step 4: If either subtree is unbalanced (helper function returned -1) or the absolute difference between left and right subtree heights is greater than 1, return -1.
Step 5: Otherwise, return the maximum of the left and right subtree heights plus 1 (the height of the current node).
Step 6: In the main `isBalanced` function, call the helper function with the root node.
Step 7: Return `true` if the helper function returned a non-negative value, `false` otherwise.

Key Insights

**Bottom-Up Balance Validation with Height Propagation:** The problem's recursive nature—every node's subtrees must be balanced—is perfectly addressed by a bottom-up (post-order) DFS. Each node is responsible for not only computing its own subtree's height but also verifying its own balance condition *after* its children have reported their status. This ensures that an imbalance deep in the tree is detected and handled before higher-level nodes attempt their own checks.
**Sentinel Value for Early Exit (Short-Circuiting):** The use of `-1` as a sentinel value is the critical optimization. Instead of a function that *only* returns height and then needing a separate boolean flag or a second pass to check balance, the `height` function itself communicates both states. If any recursive call encounters an imbalance or receives a `-1` from its children, it immediately propagates `-1` upwards. This effectively "short-circuits" the computation for that entire branch, preventing unnecessary calculations once an imbalance is confirmed.
**Single-Pass O(N) Efficiency:** By cleverly combining height calculation and balance validation into one recursive traversal, the solution processes each node exactly once. Each node performs constant work (recursive calls, difference check, max operation). This results in an optimal O(N) time complexity, where N is the number of nodes, avoiding the potential O(N log N) or O(N^2) of naive approaches that might repeatedly calculate subtree heights.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Adobe, Apple, Capgemini, Meta, Uber, Yandex.

Balanced Binary Tree | LeetCode 110

Easy

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 {
    public boolean isBalanced(TreeNode root) {
        return checkBalance(root) != -1;
    }

    private int checkBalance(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int leftHeight = checkBalance(root.left);
        if (leftHeight == -1) {
            return -1;
        }

        int rightHeight = checkBalance(root.right);
        if (rightHeight == -1) {
            return -1;
        }

        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }

        return Math.max(leftHeight, rightHeight) + 1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 110: Balanced Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

AdobeAppleCapgeminiMetaUberYandex

Stats

Difficulty:Easy

Languages:4

Companies:6

Problem #:110

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

7 steps

1
Step 1: Define a recursive helper function (e.g., `height` or `checkBalance`) that takes a tree node as input.
2
Step 2: Base case: If the node is null, return 0 (height of an empty subtree is 0).
3
Step 3: Recursively calculate the height of the left and right subtrees.
4
Step 4: If either subtree is unbalanced (helper function returned -1) or the absolute difference between left and right subtree heights is greater than 1, return -1.
5
Step 5: Otherwise, return the maximum of the left and right subtree heights plus 1 (the height of the current node).
6
Step 6: In the main `isBalanced` function, call the helper function with the root node.
7
Step 7: Return `true` if the helper function returned a non-negative value, `false` otherwise.

💡

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 the helper function visits each node in the tree exactly once. The space complexity is O(h), where h is the height of the tree, due to the recursive call stack. In the worst case (a skewed tree), h could be equal to n, resulting in O(n) space complexity. However, for balanced trees, h is log(n), leading to O(log n) space complexity.

📖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

Two Separate Height and Balanced Checks

This approach uses two separate recursive functions: one to calculate height and another to check balance. It calculates the height of left and right subtrees first and only checks balance if both subtrees are balanced.

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

Trade-offs: This approach might be slightly less efficient than the combined approach, as it makes two separate recursive traversals in some cases. However, it can improve code readability for some.

✨

Key Insights

1**Bottom-Up Balance Validation with Height Propagation:** The problem's recursive nature—every node's subtrees must be balanced—is perfectly addressed by a bottom-up (post-order) DFS. Each node is responsible for not only computing its own subtree's height but also verifying its own balance condition *after* its children have reported their status. This ensures that an imbalance deep in the tree is detected and handled before higher-level nodes attempt their own checks.
2**Sentinel Value for Early Exit (Short-Circuiting):** The use of `-1` as a sentinel value is the critical optimization. Instead of a function that *only* returns height and then needing a separate boolean flag or a second pass to check balance, the `height` function itself communicates both states. If any recursive call encounters an imbalance or receives a `-1` from its children, it immediately propagates `-1` upwards. This effectively "short-circuits" the computation for that entire branch, preventing unnecessary calculations once an imbalance is confirmed.
3**Single-Pass O(N) Efficiency:** By cleverly combining height calculation and balance validation into one recursive traversal, the solution processes each node exactly once. Each node performs constant work (recursive calls, difference check, max operation). This results in an optimal O(N) time complexity, where N is the number of nodes, avoiding the potential O(N log N) or O(N^2) of naive approaches that might repeatedly calculate subtree heights.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree. The solutions correctly handle this by returning `true` (an empty tree is considered balanced).
Edge case 2: Tree with only one node. This is also correctly handled; a single-node tree is balanced.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating height. Forgetting to add 1 to the maximum height of subtrees is a common mistake.

✗Common mistake 2: Not handling the unbalanced case efficiently. Continuing to traverse the tree after finding an unbalanced subtree leads to unnecessary computation.

✅

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: Maintaining balanced trees in databases or other data structures where efficient search and insertion are crucial. A balanced tree ensures that operations remain O(log n) rather than potentially degrading to O(n).

🏦Real-world scenario 2: Analyzing the structure of hierarchical data, such as file systems or organizational charts, to assess their efficiency and identify potential bottlenecks.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Tree. It's a great foundation for more complex problems.

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 19, 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 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

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 "Balanced Binary Tree"?

For this easy-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?