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Home/Questions/Python/Coding/Convert a sorted array into a Binary Search Tree

Convert a sorted array into a Binary Search Tree

Python/Codingeasy2 min read

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…

🤖 Analyze Your Answer
Frequency
Low
Asked at 1 company
Category
179
questions in Python/Coding
Difficulty Split
127E|24M|28H
in this category
Total Bank
1,863
across 7 categories
Asked at these companies
McKinsey

Why This Question Matters

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.

How to Approach This

Start by clearly defining the core concept being asked about. Interviewers want to see that you understand the fundamentals before diving into implementation details. Structure your answer with a definition, then explain the practical application with a concise example. The expert answer includes a code example that demonstrates the implementation pattern.

Expert Answer
408 wordsIncludes code

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.

Mechanics and Why

The algorithm operates recursively:
  • Base Case: If the low index exceeds the high index, return None (representing an empty subtree).
  • Recursive Step:
  • * Calculate the middle index: 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.

    Example Implementation

    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)

    In the interview, also mention…

    While direct BSTs are less common for raw data storage in distributed systems (which favor columnar and partitioned formats), the underlying principle of balanced tree structures is fundamental for efficient indexing and metadata management. For instance, database indexes (like B-trees in PostgreSQL) rely on similar balancing concepts for O(log N) lookups, critical for query optimizers. Even in distributed data lakes, structures like Snowflake's micro-partitions or Delta Lake's transaction log leverage organized metadata to enable efficient data pruning and version lookups, conceptually benefiting from balanced search structures.
    ⚡
    Pro Tip

    Pro-Move: Mention balance guarantee. Red Flag: O(n²) from picking wrong pivot.

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