Convert Sorted Array to Binary Search Tree - Video Solutions

Problem Overview: You receive a sorted integer array in ascending order and must convert it into a height-balanced Binary Search Tree (BST). The BST property must hold (left < root < right), and the tree height should remain balanced so that operations stay efficient.

Approach 1: Sequential BST Insertion (O(n log n) time, O(n) space)

A straightforward idea is to iterate through the sorted array and insert each value into a BST using standard BST insertion rules. Each insertion places smaller values on the left and larger values on the right. Because the array is already sorted, every new element becomes the right child of the previous node, producing a highly skewed tree that behaves like a linked list.

This approach technically constructs a valid BST, but it fails the height-balanced requirement. Time complexity becomes O(n log n) in the average case but degrades to O(n^2) when the tree becomes completely skewed. Space complexity is O(n) for storing nodes. It demonstrates BST construction basics but does not satisfy the problem constraints.

Approach 2: Recursive Middle Element as Root (O(n) time, O(log n) space)

The optimal strategy leverages the sorted property of the array. The key insight: the middle element naturally divides the array into two equal halves. Choosing the middle element as the root guarantees that the left subtree contains smaller values and the right subtree contains larger values, which preserves the BST property.

From there, recursively build the tree. Pick the middle index mid = (left + right) / 2, create a node with nums[mid], and recursively construct the left subtree from the subarray [left, mid-1] and the right subtree from [mid+1, right]. Each recursive call repeats the same logic until the subarray becomes empty.

This divide-and-conquer pattern ensures the tree remains balanced because each level splits the array roughly in half. Every element becomes a node exactly once, producing O(n) time complexity. The recursion stack depth equals the height of the balanced tree, which is O(log n) space.

This solution directly applies concepts from Divide and Conquer and builds a balanced structure using properties of a Binary Search Tree. The resulting structure is also a valid Binary Tree where subtree heights differ by at most one.

Recommended for interviews: The recursive middle-element approach is what interviewers expect. It shows you recognize how sorted data enables balanced partitioning. Mentioning the naive insertion approach demonstrates understanding of BST construction, but implementing the divide-and-conquer solution shows stronger algorithmic reasoning and leads to the optimal O(n) solution.