Convert Sorted Array to Balanced BST in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind converting a sorted array to a balanced BST?

Pick the middle element of the array as the root, then recursively do the same for left and right halves to keep the tree balanced.

beginner

Why do we choose the middle element as the root when converting a sorted array to a BST?

Choosing the middle element splits the array into two equal parts, ensuring the left and right subtrees have similar sizes, which keeps the BST balanced.

intermediate

What is the time complexity of converting a sorted array to a balanced BST?

O(n), where n is the number of elements in the array, because each element is visited once to create nodes.

intermediate

Show the recursive steps to build a balanced BST from the array [1, 2, 3, 4, 5].

1. Pick middle (3) as root. 2. Left subtree from [1, 2]: middle is 1. 3. Right child of 1 is 2. 4. Right subtree from [4, 5]: middle is 4. 5. Right child of 4 is 5. Resulting BST is balanced.

beginner

What data structure is used to represent the BST nodes in C?

A struct with an integer value and pointers to left and right child nodes.

What element should be chosen as the root when converting a sorted array to a balanced BST?

A The first element

B The last element

C The middle element

D Any random element

Choosing the middle element ensures the tree is balanced by splitting the array evenly.

What is the time complexity of building a balanced BST from a sorted array of size n?

A O(n)

B O(n^2)

C O(log n)

D O(n log n)

Each element is visited once to create nodes, so the time complexity is O(n).

Which traversal order is naturally used when converting a sorted array to a balanced BST?

A Inorder

B Preorder

C Postorder

D Level order

The recursive process picks the middle element first (root), then left and right subtrees, matching preorder traversal.

What happens if you always pick the first element as root when converting a sorted array to BST?

A The BST has duplicate nodes

B The BST is balanced

C The BST is empty

D The BST becomes skewed (like a linked list)

Picking the first element always leads to a skewed tree, losing balance.

In C, what type of variable holds the left and right children in a BST node?

A struct pointer

B float

C int

D char

Left and right children are pointers to struct nodes.

Explain step-by-step how to convert a sorted array into a balanced BST.

Think about dividing the array and building the tree from the middle.

Describe the structure of a BST node in C used for this conversion.

Remember how trees are built with nodes pointing to children.

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