How do you solve Convert Sorted Array to Binary Search Tree on LeetCode?

Convert Sorted Array to Binary Search Tree (LeetCode #108) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using the middle element of the sorted array ensures a balanced tree.

What is the brute force solution for Convert Sorted Array to Binary Search Tree?

The brute force approach for Convert Sorted Array to Binary Search Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach would involve creating a binary search tree by inserting each element of the sorted array one by one. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Convert Sorted Array to Binary Search Tree?

Always clarify whether the input array is sorted before starting. Explain your thought process clearly while coding. Consider edge cases like an empty array or an array with one element.

What are common mistakes when solving Convert Sorted Array to Binary Search Tree?

Failing to handle the base case of an empty array in recursion. Not ensuring that the tree remains balanced when inserting elements.

#108

Convert Sorted Array to Binary Search Tree

Easy

ArrayDivide and ConquerTreeBinary Search TreeBinary TreeDivide and ConquerRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach takes advantage of the sorted nature of the array. By using a divide-and-conquer strategy, we can recursively select the middle element as the root, ensuring a balanced tree.

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Algorithm

5 steps

1Step 1: If the array is empty, return null.
2Step 2: Find the middle index of the array.
3Step 3: Create a new TreeNode with the middle element as the root.
4Step 4: Recursively build the left subtree using the left half of the array.
5Step 5: Recursively build the right subtree using the right half of the array.

solution.py15 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def sorted_array_to_bst(nums):
9    if not nums:
10        return None
11    mid = len(nums) // 2
12    root = TreeNode(nums[mid])
13    root.left = sorted_array_to_bst(nums[:mid])
14    root.right = sorted_array_to_bst(nums[mid+1:])
15    return root

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Complexity note: The time complexity is O(n) because we visit each element once to construct the tree. The space complexity is O(n) due to the recursion stack in the worst case.

1Using the middle element of the sorted array ensures a balanced tree.
2Recursion can simplify the process of building the tree from the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.