Convert Sorted Array to Balanced BST in DSA C - Time & Space Complexity

Time Complexity: Convert Sorted Array to Balanced BST

O(n)

Understanding Time Complexity

We want to know how the time needed to build a balanced binary search tree changes as the input array grows.

Specifically, how does the work increase when the sorted array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct TreeNode {
    int val;
    struct TreeNode *left, *right;
};

struct TreeNode* sortedArrayToBST(int* nums, int left, int right) {
    if (left > right) return NULL;
    int mid = left + (right - left) / 2;
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = nums[mid];
    root->left = sortedArrayToBST(nums, left, mid - 1);
    root->right = sortedArrayToBST(nums, mid + 1, right);
    return root;
}
    

This code builds a balanced binary search tree by choosing the middle element as root and recursively doing the same for left and right halves.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls splitting the array and creating tree nodes.
• How many times: Each element is visited once to create a node, so total calls equal the number of elements.

How Execution Grows With Input

Each element in the array leads to one node creation and recursive calls split the array in halves.

Input Size (n)	Approx. Operations
10	About 10 node creations and recursive calls
100	About 100 node creations and recursive calls
1000	About 1000 node creations and recursive calls

Pattern observation: The work grows roughly in direct proportion to the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to build the tree grows linearly with the number of elements in the array.

Common Mistake

[X] Wrong: "Because the function calls itself twice, the time complexity is O(2^n)."

[OK] Correct: Each element is processed only once to create a node, so calls do not multiply exponentially but split the array into smaller parts.

Interview Connect

Understanding this time complexity helps you explain how recursive tree building works efficiently and shows you can analyze divide-and-conquer algorithms clearly.

Self-Check

"What if we changed the input from a sorted array to an unsorted array? How would the time complexity change?"

Practice

1. What is the main goal when converting a sorted array to a balanced Binary Search Tree (BST)?

easy

A. To create a linked list from the array

B. To create a BST with minimal height

C. To reverse the array elements

D. To sort the array again

Solution

1. Step 1: Understand the problem goal

   The goal is to convert a sorted array into a BST that is balanced, meaning the tree height is minimized for efficient operations.

2. Step 2: Identify balanced BST property

   A balanced BST has roughly equal nodes on left and right subtrees, minimizing height and ensuring faster search.

3. Final Answer:

   To create a BST with minimal height -> Option B

4. Quick Check:

   Balanced BST = minimal height [OK]

Hint: Balanced BST means minimal height tree from sorted array [OK]

Common Mistakes:

• Thinking the array needs to be reversed
• Confusing BST with linked list
• Ignoring the balance requirement

2. Which of the following is the correct base case condition in the recursive function to convert a sorted array to a balanced BST?

easy

A. if (start == end) return NULL;

B. if (start < end) return NULL;

C. if (start > end) return NULL;

D. if (start == 0) return NULL;

Solution

1. Step 1: Understand recursion base case

   When start index is greater than end index, it means no elements remain to process, so return NULL (no node).

2. Step 2: Check each option

   if (start > end) return NULL; correctly checks if start > end to stop recursion. Other options do not correctly represent the base case.

3. Final Answer:

   if (start > end) return NULL; -> Option C

4. Quick Check:

   Base case = start > end [OK]

Hint: Base case stops when start index passes end index [OK]

Common Mistakes:

• Using start == end as base case incorrectly
• Returning NULL too early
• Confusing start and end indices

3. Given the sorted array [1, 2, 3, 4, 5], what is the root value of the balanced BST created by the standard algorithm?

medium

A. 3

B. 1

C. 5

D. 4

Solution

1. Step 1: Find middle element of array

   The middle element of [1, 2, 3, 4, 5] is at index 2 (0-based), which is 3.

2. Step 2: Middle element becomes root

   In balanced BST from sorted array, middle element is root to balance left and right subtrees.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Middle element = root = 3 [OK]

Hint: Root is middle element of sorted array [OK]

Common Mistakes:

• Choosing first or last element as root
• Off-by-one error in middle index calculation
• Confusing array indices

4. Identify the error in this recursive function snippet for converting a sorted array to a balanced BST:

TreeNode* sortedArrayToBST(int arr[], int start, int end) {
  if (start >= end) return NULL;
  int mid = (start + end) / 2;
  TreeNode* root = new TreeNode(arr[mid]);
  root->left = sortedArrayToBST(arr, start, mid - 1);
  root->right = sortedArrayToBST(arr, mid + 1, end);
  return root;
}

medium

A. Right subtree call should use mid instead of mid + 1

B. Mid calculation should be (start + end + 1) / 2

C. Left subtree call should use mid instead of mid - 1

D. Base case condition should be start > end, not start >= end

Solution

1. Step 1: Analyze base case condition

   The base case should stop recursion when start > end, because when start == end, there is still one element to process.

2. Step 2: Check recursive calls

   Left subtree uses start to mid-1, right subtree uses mid+1 to end, which is correct.

3. Final Answer:

   Base case condition should be start > end, not start >= end -> Option D

4. Quick Check:

   Base case = start > end [OK]

Hint: Base case excludes start == end to include single element nodes [OK]

Common Mistakes:

• Stopping recursion too early with start >= end
• Incorrect mid index calculation
• Wrong subtree index ranges

5. You have a sorted array [2, 4, 6, 8, 10, 12, 14]. After converting it to a balanced BST, what is the inorder traversal of the BST?

hard

A. [2, 4, 6, 8, 10, 12, 14]

B. [8, 4, 2, 6, 12, 10, 14]

C. [14, 12, 10, 8, 6, 4, 2]

D. [6, 4, 2, 8, 10, 12, 14]

Solution

1. Step 1: Understand inorder traversal of BST

   Inorder traversal of a BST always returns the sorted order of elements.

2. Step 2: Check given array and traversal

   The array is already sorted. After conversion to balanced BST, inorder traversal must return the same sorted array.

3. Final Answer:

   [2, 4, 6, 8, 10, 12, 14] -> Option A

4. Quick Check:

   Inorder traversal = sorted array [OK]

Hint: Inorder traversal of BST returns sorted array [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Expecting different order after conversion
• Misunderstanding BST traversal properties