# Convert Sorted Array to Binary Search Tree Problem on LeetCode in Python

## Introduction

Binary search trees (BST) are a fundamental data structure in computer science that allows for efficient storage and retrieval of elements. The Convert Sorted Array to Binary Search Tree problem on LeetCode challenges us to create a balanced BST from a given sorted array. In this article, we’ll dive into various ways to approach this problem using Python, and understand their efficiencies and trade-offs.

1. Introduction
2. Understanding the Problem Statement
3. Recursive Approach to Building a Balanced BST
4. Analyzing Time and Space Complexities
5. Conclusion

## Understanding the Problem Statement

The Convert Sorted Array to Binary Search Tree problem is defined as follows:

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

Example:

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]

The task is to create a balanced BST, which means the depths of the two subtrees of any node in the tree should not differ by more than one.

## Recursive Approach to Building a Balanced BST

A natural approach to this problem is to build the tree recursively. Given a sorted array, the middle element can be chosen as the root of the BST. The elements to the left of the middle element make up the left subtree, and the elements to the right make up the right subtree. We can apply this logic recursively to build the BST. By selecting the middle element as the root, we are ensuring that the number of elements in the left and right subtrees are as equal as possible, which is the condition for the BST to be balanced.

Here’s what the Python code looks like:

class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right

class Solution:
def sortedArrayToBST(self, nums):
# Base case
if not nums:
return None

# Find the middle element
mid = len(nums) // 2

# Root's value is the middle element
root = TreeNode(nums[mid])

# Recursively construct the left and right subtrees
root.left = self.sortedArrayToBST(nums[:mid]) # Elements before mid
root.right = self.sortedArrayToBST(nums[mid+1:]) # Elements after mid

return root

## Analyzing Time and Space Complexities

### Time Complexity

The time complexity of this approach is O(n), where n is the number of elements in the array. This is because we visit each element exactly once. The total number of operations is proportional to the number of elements in the array.

### Space Complexity

The space complexity is O(log n) for a balanced tree, but in the worst case, it can be O(n). This space is used by the call stack during the recursion. In the average case where the tree is balanced, the height of the tree is log(n), and that many recursive calls are on the stack at once. In the worst case where the tree is unbalanced (degenerate to a linked list), there will be n recursive calls on the stack.

## Conclusion

The Convert Sorted Array to Binary Search Tree problem on LeetCode is a classical application of binary trees. In this article, we investigated a recursive approach for constructing a balanced BST from a sorted array. This recursive approach is efficient and concise. Understanding the construction of binary search trees is essential as BSTs form the basis for several advanced data structures and algorithms.