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 to build a balanced binary search tree grows as the input array size increases.

How does the number of steps change when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

TreeNode* sortedArrayToBST(std::vector<int>& nums, int left, int right) {
    if (left > right) return nullptr;
    int mid = left + (right - left) / 2;
    TreeNode* root = new TreeNode(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 picking the middle element as root and recursively doing the same for left and right parts.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls splitting the array and creating nodes.
• How many times: Each element is visited once to create a node, so recursion happens about n times for n elements.

How Execution Grows With Input

Each element leads to one node creation and two recursive calls on smaller parts.

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 number of operations 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 exponential like O(2^n)."

[OK] Correct: Each element is processed only once to create a node, so the total work is proportional to n, not exponential.

Interview Connect

Understanding this helps you explain how recursion can efficiently build balanced trees, a common task in coding interviews.

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 BST with minimum height

B. To create a BST with maximum height

C. To create a BST with all nodes having two children

D. To create a BST with nodes in reverse order

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.

2. Step 2: Reason about balanced BST

   A balanced BST has minimum height to ensure efficient search operations.

3. Final Answer:

   To create a BST with minimum height -> Option A

4. Quick Check:

   Balanced BST = Minimum height [OK]

Hint: Pick the option that minimizes tree height for balance [OK]

Common Mistakes:

• Thinking maximum height means balanced
• Assuming all nodes must have two children
• Confusing order of nodes with balance

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 nullptr;

B. if (start == end) return nullptr;

C. if (start < end) return nullptr;

D. if (start >= end) return nullptr;

Solution

1. Step 1: Identify base case for recursion

   The recursion should stop when the start index passes the end index, meaning no elements left.

2. Step 2: Match condition to code

   The condition if (start > end) return nullptr; correctly handles this base case.

3. Final Answer:

   if (start > end) return nullptr; -> Option A

4. Quick Check:

   Base case = start > end [OK]

Hint: Stop recursion when start index exceeds end index [OK]

Common Mistakes:

• Using start == end as base case incorrectly
• Using start < end which never stops recursion
• Using start >= end which excludes valid single element

3. Given the sorted array [2, 4, 6, 8, 10, 12, 14], which of the following is the correct inorder traversal of the balanced BST created by the standard conversion method?

easy

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

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

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

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

Solution

1. Step 1: Understand inorder traversal of BST

   Inorder traversal of a BST always returns the nodes in sorted order.

2. Step 2: Check given array and traversal

   The array is sorted. The balanced BST created from it will have inorder traversal same as the original array.

3. Final Answer:

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

4. Quick Check:

   Inorder traversal = sorted array [OK]

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Assuming balanced BST changes inorder order
• Selecting reversed or mixed order arrays

4. What is the output BST preorder traversal for the sorted array [1, 2, 3, 4, 5] when converted to a balanced BST?

medium

A. [3, 2, 1, 5, 4]

B. [3, 1, 2, 4, 5]

C. [3, 2, 1, 4, 5]

D. [3, 1, 4, 2, 5]

Solution

1. Step 1: Build balanced BST from array

   Middle element is 3 (index 2), root node. Left subtree from [1,2], right subtree from [4,5].

2. Step 2: Construct preorder traversal

   Preorder = root, left subtree preorder, right subtree preorder.
   Left subtree indices [0,1], mid=(0+1)/2=0 root=1, right child from [1,1] root=2.
   Right subtree indices [3,4], mid=(3+4)/2=3 root=4, right child from [4,4] root=5.
   Preorder: 3, 1, 2, 4, 5.

3. Final Answer:

   [3, 1, 2, 4, 5] -> Option B

4. Quick Check:

   Preorder = root-left-right = [3,1,2,4,5] [OK]

Hint: Pick middle as root, preorder = root-left-right [OK]

Common Mistakes:

• Choosing wrong middle element for root
• Mixing preorder with inorder or postorder
• Incorrect subtree construction

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

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

medium

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

B. Mid calculation should be mid = (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 stops recursion when no elements remain. Using start >= end excludes the case when start == end, missing single element nodes.

2. Step 2: Correct base case for single element

   Base case should be start > end to allow processing single elements when start == end.

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 must allow single element processing [OK]

Common Mistakes:

• Using start >= end excludes valid nodes
• Changing mid calculation unnecessarily
• Incorrect subtree index ranges