Convert Sorted Array to Balanced BST in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle index of the array segment.

DSA C

int mid = (start + [1]) / 2;

Drag options to blanks, or click blank then click option'

Aend

Bstart

C0

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using start instead of end causes incorrect middle calculation.

Using size instead of end leads to wrong index.

The middle index is calculated by adding start and end indices and dividing by 2.

2fill in blank

medium

Complete the code to create a new tree node with the middle element.

DSA C

TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
node->val = [1][mid];

Drag options to blanks, or click blank then click option'

Aarr

Bnode

Cmid

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using node instead of arr causes wrong assignment.

Using mid as a value instead of index is incorrect.

The node value is assigned from the array element at the middle index.

3fill in blank

hard

Fix the error in the recursive call to build the left subtree.

DSA C

node->left = sortedArrayToBST(arr, [1], mid - 1);

Drag options to blanks, or click blank then click option'

A0

Bmid

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid instead of start causes wrong segment.

Using end instead of start leads to errors.

The left subtree uses the start index to mid - 1 segment.

4fill in blank

hard

Fill both blanks to build the right subtree correctly.

DSA C

node->right = sortedArrayToBST(arr, [1], [2]);

Drag options to blanks, or click blank then click option'

Amid + 1

Bstart

Cend

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using start instead of mid + 1 causes overlap.

Using mid instead of mid + 1 misses elements.

The right subtree covers from mid + 1 to end indices.

5fill in blank

hard

Fill all three blanks to complete the base case and recursive call.

DSA C

if ([1] > [2]) return NULL;
int mid = (start + end) / 2;
TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
node->val = arr[mid];
node->left = sortedArrayToBST(arr, start, mid - 1);
node->right = sortedArrayToBST(arr, [3], end);
return node;

Drag options to blanks, or click blank then click option'

Astart

Bend

Cmid + 1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping start and end in base case causes infinite recursion.

Using mid instead of mid + 1 for right subtree misses elements.

The base case checks if start is greater than end. The right subtree starts from mid + 1.

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