Convert Sorted Array to Balanced BST in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Balanced BST Inorder Traversal

Given the following JavaScript code that converts a sorted array to a balanced BST and then prints its inorder traversal, what is the output?

DSA Javascript

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function sortedArrayToBST(nums) {
  if (!nums.length) return null;
  const mid = Math.floor(nums.length / 2);
  const root = new TreeNode(nums[mid]);
  root.left = sortedArrayToBST(nums.slice(0, mid));
  root.right = sortedArrayToBST(nums.slice(mid + 1));
  return root;
}

function inorderTraversal(root) {
  if (!root) return [];
  return [...inorderTraversal(root.left), root.val, ...inorderTraversal(root.right)];
}

const nums = [1, 2, 3, 4, 5, 6, 7];
const bstRoot = sortedArrayToBST(nums);
console.log(inorderTraversal(bstRoot).join(' -> ') + ' -> null');

A 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null

B 4 -> 2 -> 1 -> 3 -> 6 -> 5 -> 7 -> null

C 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> null

D null

💡 Hint

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

The function builds a balanced BST by choosing the middle element as root recursively. Inorder traversal of a BST always returns the sorted array. So the output matches the original sorted array.

🧠 Conceptual

intermediate

Height of Balanced BST from Sorted Array

If you convert a sorted array of length 15 into a balanced BST using the middle element as root recursively, what will be the height of the resulting BST?

A 4

B 3

C 5

D 15

💡 Hint

Height of balanced BST is about log2(n+1).

A balanced BST built from a sorted array has height approximately log base 2 of (n+1). For n=15, log2(16) = 4.

🔧 Debug

advanced

Identify the Bug in BST Construction

What is the output of this code snippet? Identify if there is a bug and what it causes.

DSA Javascript

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function sortedArrayToBST(nums) {
  if (nums.length === 0) return null;
  const mid = Math.floor(nums.length / 2);
  const root = new TreeNode(nums[mid]);
  root.left = sortedArrayToBST(nums.slice(0, mid));
  root.right = sortedArrayToBST(nums.slice(mid + 1));
  return root;
}

function preorderTraversal(root) {
  if (!root) return [];
  return [root.val, ...preorderTraversal(root.left), ...preorderTraversal(root.right)];
}

const nums = [1, 2, 3, 4, 5];
const bstRoot = sortedArrayToBST(nums);
console.log(preorderTraversal(bstRoot).join(' -> ') + ' -> null');

A 3 -> 1 -> 2 -> 5 -> 4 -> null

B 4 -> 2 -> 1 -> 3 -> 5 -> null (with off-by-one bug)

C 4 -> 2 -> 1 -> 3 -> 5 -> null

D TypeError due to invalid slice indices

💡 Hint

Check how mid index is calculated and how slice works.

Using Math.ceil for mid causes the root to be chosen incorrectly, leading to an unbalanced tree and incorrect preorder traversal output.

🚀 Application

advanced

Number of Nodes in Balanced BST

If you build a balanced BST from a sorted array of length 31, how many nodes will the left subtree of the root have?

A 30

B 16

C 15

D 31

💡 Hint

The root is the middle element, left subtree has elements before it.

For 31 elements, middle index is 15 (0-based), so left subtree has 15 nodes (indices 0 to 14).

❓ Predict Output

expert

Output of Level Order Traversal of Balanced BST

What is the output of the level order traversal of the balanced BST created from the sorted array [10, 20, 30, 40, 50, 60, 70]?

DSA Javascript

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function sortedArrayToBST(nums) {
  if (!nums.length) return null;
  const mid = Math.floor(nums.length / 2);
  const root = new TreeNode(nums[mid]);
  root.left = sortedArrayToBST(nums.slice(0, mid));
  root.right = sortedArrayToBST(nums.slice(mid + 1));
  return root;
}

function levelOrderTraversal(root) {
  if (!root) return [];
  const queue = [root];
  const result = [];
  while (queue.length) {
    const node = queue.shift();
    result.push(node.val);
    if (node.left) queue.push(node.left);
    if (node.right) queue.push(node.right);
  }
  return result;
}

const nums = [10, 20, 30, 40, 50, 60, 70];
const bstRoot = sortedArrayToBST(nums);
console.log(levelOrderTraversal(bstRoot).join(' -> ') + ' -> null');

A 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> null

B null

C 70 -> 60 -> 50 -> 40 -> 30 -> 20 -> 10 -> null

D 40 -> 20 -> 60 -> 10 -> 30 -> 50 -> 70 -> null

💡 Hint

Level order visits nodes level by level from root.

The middle element 40 is root, left subtree root 20, right subtree root 60. Level order visits 40, then 20 and 60, then their 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 reverse the array elements

C. To create a BST with minimal height for efficient search

D. To sort the array again

Solution

1. Step 1: Understand the problem goal

   The goal is to build a BST that is balanced, meaning the tree height is minimized for faster search operations.

2. Step 2: Recognize the importance of balance

   A balanced BST ensures operations like search, insert, and delete run efficiently, typically in O(log n) time.

3. Final Answer:

   To create a BST with minimal height for efficient search -> Option C

4. Quick Check:

   Balanced BST = minimal height = efficient search [OK]

Hint: Pick the option about balanced tree and efficient search [OK]

Common Mistakes:

• Thinking the goal is to reverse or sort the array again
• Confusing BST with linked list
• Ignoring the balance aspect of the tree

2. Which of the following is the correct way to find the middle index of an array arr in JavaScript for building a balanced BST?

easy

A. const mid = Math.floor(arr.length / 2);

B. const mid = arr.length / 2;

C. const mid = Math.ceil(arr.length / 2);

D. const mid = arr.length - 1;

Solution

1. Step 1: Understand array indexing

   Array indices start at 0, so to get the middle index, we divide length by 2 and take the floor to get an integer index.

2. Step 2: Use Math.floor for integer index

   Using Math.floor(arr.length / 2) gives the correct middle index for both even and odd lengths.

3. Final Answer:

   const mid = Math.floor(arr.length / 2); -> Option A

4. Quick Check:

   Middle index = floor(length/2) [OK]

Hint: Use Math.floor to get integer middle index [OK]

Common Mistakes:

• Using division without floor, resulting in float index
• Using Math.ceil which may pick wrong middle
• Using last index instead of middle

3. What will be the root node value of the balanced BST created from the sorted array [1, 3, 5, 7, 9]?

medium

A. 3

B. 5

C. 7

D. 9

Solution

1. Step 1: Find the middle element of the array

   The array length is 5, middle index = floor(5/2) = 2, so middle element is arr[2] = 5.

2. Step 2: Middle element becomes root

   In balanced BST from sorted array, the middle element is chosen as root to keep tree balanced.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Middle element = root = 5 [OK]

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

Common Mistakes:

• Picking first or last element as root
• Using ceil instead of floor for middle index
• Confusing array index with value

4. Consider this code snippet to convert a sorted array to a balanced BST:

function sortedArrayToBST(arr) {
  if (!arr.length) return null;
  const mid = Math.floor(arr.length / 2);
  const root = { val: arr[mid], left: null, right: null };
  root.left = sortedArrayToBST(arr.slice(0, mid));
  root.right = sortedArrayToBST(arr.slice(mid));
  return root;
}

What is the bug in this code?

medium

A. The right subtree slice should start from mid + 1, not mid

B. The base case should check for arr === null instead of length

C. The mid index calculation is incorrect

D. The root node is not created properly

Solution

1. Step 1: Analyze the slicing for right subtree

   The right subtree should exclude the middle element already used as root, so slice should start from mid + 1.

2. Step 2: Identify the bug

   Using arr.slice(mid) includes the middle element again in the right subtree, causing duplicate nodes and incorrect tree.

3. Final Answer:

   The right subtree slice should start from mid + 1, not mid -> Option A

4. Quick Check:

   Right subtree slice = arr.slice(mid + 1) [OK]

Hint: Right subtree excludes middle element: slice from mid+1 [OK]

Common Mistakes:

• Including middle element in both subtrees
• Wrong base case condition
• Incorrect mid calculation

5. Given the sorted array [-10, -3, 0, 5, 9], what is the preorder traversal (root-left-right) of the balanced BST created by the standard algorithm?

hard

A. [0, -10, -3, 5, 9]

B. [-3, -10, 0, 5, 9]

C. [5, 0, -10, -3, 9]

D. [0, -3, -10, 9, 5]

Solution

1. Step 1: Build the balanced BST

   Middle index = Math.floor(5/2) = 2, root = 0. Left subarray [-10, -3], length=2, mid=Math.floor(2/2)=1, left child = -3 (left of -3: [-10] mid=0 value -10; right of -3: empty). Right subarray [5, 9], length=2, mid=1, right child = 9 (left of 9: [5]; right of 9: empty).

2. Step 2: Perform preorder traversal (root-left-right)

   0, then left subtree: -3 then -10, then right subtree: 9 then 5. Result: [0, -3, -10, 9, 5].

3. Final Answer:

   [0, -3, -10, 9, 5] -> Option D

4. Quick Check:

   Preorder = root-left-right = [0, -3, -10, 9, 5] [OK]

Hint: Preorder = root, then left subtree, then right subtree [OK]

Common Mistakes:

• Wrong middle element selection for subtrees
• Confusing preorder with inorder or postorder
• Incorrect subtree construction