What is the time complexity of Convert Sorted Array to Binary Search Tree?

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

What is the space complexity of Convert Sorted Array to Binary Search Tree?

The space complexity is O(log n).

What data structures are used to solve Convert Sorted Array to Binary Search Tree?

This problem uses Array, Divide and Conquer, Tree, Binary Search Tree, Binary Tree. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Convert Sorted Array to Binary Search Tree?

**Midpoint as the Balancing Pivot**: The fundamental insight is that selecting the middle element of a *sorted* array to be the root node naturally leads to a height-balanced BST. This choice evenly distributes the remaining elements to the left and right subtrees, minimizing the height difference and preventing skew. **Recursive Divide and Conquer**: The problem structure perfectly aligns with a recursive approach. Building the complete BST can be elegantly broken down into creating a root, then recursively solving the identical smaller problem for the left sub-array (to build the left child), and similarly for the right sub-array (for the right child). This pattern ensures that each part of the tree adheres to the balance property.

Which companies ask LeetCode 108. Convert Sorted Array to Binary Search Tree in interviews?

This problem is frequently asked at Adobe, Airbnb, Apple, Meta, Samsung.

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Convert Sorted Array to Binary Search Tree - LeetCode 108 Solution

Q: How to solve Convert Sorted Array to Binary Search Tree?

The provided solution masterfully employs recursion, specifically a divide-and-conquer strategy, to construct the height-balanced BST. The central idea revolves around the fact that for a sorted array, the *middle element* is the optimal choice for the root of a (sub)tree. Why? Because it naturally partitions the remaining elements into two groups of roughly equal size: those smaller than it (for the left subtree) and those larger than it (for the right subtree). The algorithm kicks off with a base case: if the input array `nums` is empty, there's no node to create, so it returns `None`. Otherwise, it finds the middle index (`len(nums) // 2`), creates a `TreeNode` with the value at that `mid` index, and designates it as the current root. The magic then happens with two recursive calls: `root.left` is built by calling `sortedArrayToBST` on the left half of the array (`nums[:mid]`), and `root.right` is built from the right half (`nums[mid+1:]`). By consistently choosing the middle element as the root at every level of recursion, the tree naturally grows with its branches having approximately the same number of nodes. This consistent 'even split' is precisely what guarantees the resulting tree will be height-balanced.

Convert Sorted Array to Binary Search Tree - Complete Solution Guide

Convert Sorted Array to Binary Search Tree is LeetCode problem 108, 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 an integer array nums where the elements are sorted in ascending order , convert it to a height-balanced binary search tree . Example 1: Input: nums = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: [0,-10,5,null,-3,null,9] is also accepted: Example 2: Input: nums = [1,3] Output: [3,1] Explanation: [1,null,3] and [3,1] are both height-balanced BSTs. Constraints: 1 <= nums.length <= 10 4 -10 4 <= nums[i] <= 10 4 nums is sorted in a strictly increasing order.

Detailed Explanation

Our mission here is to transform a simple array of numbers, which is already conveniently sorted in ascending order, into a specific tree structure: a Binary Search Tree (BST) that is also 'height-balanced'. The 'sorted' input is a huge gift, but the 'height-balanced' part is the real constraint. It means that for every single node in our new tree, the heights of its left and right subtrees can differ by at most one. This prevents the tree from degenerating into a linked list and ensures efficient search operations. Consider the example `nums = [-10,-3,0,5,9]`. If we just picked `-10` as the root and built the tree, it would be heavily skewed right, making it effectively a linked list and far from height-balanced. The sorted nature of the array guides us: in a BST, all values to the left of a node are smaller, and all to the right are larger. To achieve height-balance, we intuitively want the root to split the remaining elements as evenly as possible. For our example, `0` is the perfect candidate: `[-10,-3]` become the left candidates, and `[5,9]` the right. This symmetry is the secret to building a compact, balanced tree. The problem statement's allowance for multiple valid outputs for simple cases like `[1,3]` (e.g., `[3,1]` or `[1,null,3]`) reinforces that any strategy which consistently produces *a* height-balanced BST, adhering to both BST rules and the height constraint, is perfectly acceptable. This frees us from finding a single canonical output, letting us focus on the structural properties.

Solution Approach

The provided solution masterfully employs recursion, specifically a divide-and-conquer strategy, to construct the height-balanced BST. The central idea revolves around the fact that for a sorted array, the *middle element* is the optimal choice for the root of a (sub)tree. Why? Because it naturally partitions the remaining elements into two groups of roughly equal size: those smaller than it (for the left subtree) and those larger than it (for the right subtree). The algorithm kicks off with a base case: if the input array `nums` is empty, there's no node to create, so it returns `None`. Otherwise, it finds the middle index (`len(nums) // 2`), creates a `TreeNode` with the value at that `mid` index, and designates it as the current root. The magic then happens with two recursive calls: `root.left` is built by calling `sortedArrayToBST` on the left half of the array (`nums[:mid]`), and `root.right` is built from the right half (`nums[mid+1:]`). By consistently choosing the middle element as the root at every level of recursion, the tree naturally grows with its branches having approximately the same number of nodes. This consistent 'even split' is precisely what guarantees the resulting tree will be height-balanced.

Step-by-Step Algorithm

Step 1: Check for the base case: If the input array `nums` is empty, return `null` (or `nullptr` in C++).
Step 2: Find the middle index `mid` of the array.
Step 3: Create a new `TreeNode` with the value at `nums[mid]` as the root.
Step 4: Recursively call the function on the left subarray (`nums[:mid]` in Python, `nums[left, mid - 1]` in Java/C++/C) to construct the left subtree and assign it to `root.left`.
Step 5: Recursively call the function on the right subarray (`nums[mid+1:]` in Python, `nums[mid + 1, right]` in Java/C++/C) to construct the right subtree and assign it to `root.right`.
Step 6: Return the `root` node.

Key Insights

**Midpoint as the Balancing Pivot**: The fundamental insight is that selecting the middle element of a *sorted* array to be the root node naturally leads to a height-balanced BST. This choice evenly distributes the remaining elements to the left and right subtrees, minimizing the height difference and preventing skew.
**Recursive Divide and Conquer**: The problem structure perfectly aligns with a recursive approach. Building the complete BST can be elegantly broken down into creating a root, then recursively solving the identical smaller problem for the left sub-array (to build the left child), and similarly for the right sub-array (for the right child). This pattern ensures that each part of the tree adheres to the balance property.
**Leveraging the Sorted Property Directly**: The fact that the input array `nums` is already sorted is not just a hint, but the core enabler for this efficient solution. It means we don't need to perform any sorting or complex partitioning to satisfy the BST property; simply picking the middle element provides a ready-made separation into 'lesser' and 'greater' values suitable for left and right children.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(log n)

Topics

This problem involves: Array, Divide and Conquer, Tree, Binary Search Tree, Binary Tree.

Companies

Asked at: Adobe, Airbnb, Apple, Meta, Samsung.

Convert Sorted Array to Binary Search Tree | LeetCode 108

Easy

Array Divide and Conquer Tree Binary Search Tree Binary Tree

Time: O(n)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 TreeNode sortedArrayToBST(int[] nums) {
        return helper(nums, 0, nums.length - 1);
    }

    private TreeNode helper(int[] nums, int left, int right) {
        if (left > right) {
            return null;
        }

        int mid = left + (right - left) / 2;
        TreeNode root = new TreeNode(nums[mid]);

        root.left = helper(nums, left, mid - 1);
        root.right = helper(nums, mid + 1, right);

        return root;
    }
}

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Problem Info

Description

Problem 108: Convert Sorted Array to Binary Search Tree - Detailed explanation and solution approaches.

Topics

Array Divide and Conquer Tree Binary Search Tree Binary Tree

Companies

AdobeAirbnbAppleMetaSamsung

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:108

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

6 steps

1
Step 1: Check for the base case: If the input array `nums` is empty, return `null` (or `nullptr` in C++).
2
Step 2: Find the middle index `mid` of the array.
3
Step 3: Create a new `TreeNode` with the value at `nums[mid]` as the root.
4
Step 4: Recursively call the function on the left subarray (`nums[:mid]` in Python, `nums[left, mid - 1]` in Java/C++/C) to construct the left subtree and assign it to `root.left`.
5
Step 5: Recursively call the function on the right subarray (`nums[mid+1:]` in Python, `nums[mid + 1, right]` in Java/C++/C) to construct the right subtree and assign it to `root.right`.
6
Step 6: Return the `root` node.

💡

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(log n)

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is O(n) because each element in the array is visited and processed exactly once during the recursive calls. The space complexity is dominated by the recursion depth. In a balanced tree, this depth is O(log n). However, in the worst case (e.g., an already sorted array), it can become O(n), resulting in a skewed 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 Approach

Instead of recursion, an iterative approach using a stack or queue can be used to construct the BST. This avoids the potential stack overflow issues associated with deeply recursive calls.

Time: O(n)Space: O(log n) (average), O(n) (worst)

Trade-offs: While both have the same time and average space complexity, the iterative approach might be slightly more efficient in certain cases and safer for very large arrays to prevent stack overflow exceptions.

✨

Key Insights

1**Midpoint as the Balancing Pivot**: The fundamental insight is that selecting the middle element of a *sorted* array to be the root node naturally leads to a height-balanced BST. This choice evenly distributes the remaining elements to the left and right subtrees, minimizing the height difference and preventing skew.
2**Recursive Divide and Conquer**: The problem structure perfectly aligns with a recursive approach. Building the complete BST can be elegantly broken down into creating a root, then recursively solving the identical smaller problem for the left sub-array (to build the left child), and similarly for the right sub-array (for the right child). This pattern ensures that each part of the tree adheres to the balance property.
3**Leveraging the Sorted Property Directly**: The fact that the input array `nums` is already sorted is not just a hint, but the core enabler for this efficient solution. It means we don't need to perform any sorting or complex partitioning to satisfy the BST property; simply picking the middle element provides a ready-made separation into 'lesser' and 'greater' values suitable for left and right children.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty array: The code handles this by returning `null` or `nullptr` at the start, preventing errors.
Edge case 2: Array with one element: The code correctly creates a BST with a single node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the middle index, leading to unbalanced trees or incorrect node values. Using `(left + right) / 2` might lead to integer overflow for very large arrays; `left + (right - left) / 2` is safer.

✗Common mistake 2: Forgetting the base case of an empty array, leading to infinite recursion and stack overflow errors.

✅

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: Efficiently storing and searching sorted data, such as sorted lists of products, user IDs, or sensor readings. BSTs provide O(log n) search time on average.

🏦Real-world scenario 2: Implementing self-balancing search trees (like AVL trees or red-black trees) for data structures where efficient search, insertion, and deletion are crucial, such as database indexes.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 17, 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

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

@CodeNinjaMay 29

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadMay 22

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

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 "Convert Sorted Array to Binary Search Tree"?

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

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