Reviewed by Aditya Kumar · Last reviewed 2026-03-24
To convert a sorted array into a Binary Search Tree (BST), the most efficient and common approach is to recursively select the middle element of the array as the root, then build its left subtree from…
This easy-level Python/Coding question appears frequently in data engineering interviews at companies like McKinsey. While less common, it tests deeper understanding that distinguishes strong candidates.
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To convert a sorted array into a Binary Search Tree (BST), the most efficient and common approach is to recursively select the middle element of the array as the root, then build its left subtree from the left half of the array and its right subtree from the right half. This method inherently constructs a height-balanced BST.
low index exceeds the high index, return None (representing an empty subtree).mid = (low + high) // 2.
* Create a TreeNode with nums[mid] as its value. This node becomes the root of the current subtree.
* Recursively call the function for the left half of the array (low to mid - 1) to build the left child of the current root.
* Recursively call the function for the right half of the array (mid + 1 to high) to build the right child of the current root.
* Return the current root.
This process ensures a balanced BST because at each step, we divide the remaining elements as evenly as possible between the left and right subtrees. A balanced BST is crucial as it guarantees O(log N) time complexity for search, insertion, and deletion operations, preventing worst-case O(N) scenarios found in skewed BSTs. The time complexity for building the tree is O(N) since each element is processed exactly once. The space complexity is O(log N) due to the recursion stack depth for a balanced tree.
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def sortedArrayToBST(nums):
def build(low, high):
if low > high:
return None
mid = (low + high) // 2
root = TreeNode(nums[mid])
root.left = build(low, mid - 1)
root.right = build(mid + 1, high)
return root
return build(0, len(nums) - 1)
Pro-Move: Mention balance guarantee. Red Flag: O(n²) from picking wrong pivot.
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