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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

const mid = Math.[1]((start + end) / 2);

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Aabs

Bfloor

Cround

Dceil

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.ceil causes index to go out of range.

Using Math.round can pick wrong middle for even length arrays.

We use Math.floor to get the middle index as an integer by rounding down.

2fill in blank

medium

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

DSA Javascript

const node = new TreeNode([1][mid]);

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Aarr

Barray

Cnums

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variable names like 'arr' or 'list'.

The input array is named 'nums', so we use nums[mid] to get the middle element.

3fill in blank

hard

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

DSA Javascript

node.left = buildBST(nums, start, [1] - 1);

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Amid

Bstart

Cend

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'mid + 1' causes overlap with right subtree.

Using 'end' or 'start' incorrectly limits the range.

We use 'mid - 1' to build the left subtree from start to just before mid.

4fill in blank

hard

Fill both blanks to build the right subtree correctly.

DSA Javascript

node.right = buildBST(nums, [1], [2]);

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Amid + 1

Bmid - 1

Cend

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 as start index for right subtree.

Using start as end index causes wrong range.

Right subtree starts from mid + 1 to end of the array segment.

5fill in blank

hard

Fill all three blanks to complete the main function that calls the helper.

DSA Javascript

function sortedArrayToBST([1]) {
  function buildBST([2], start, end) {
    if (start > end) return null;
    const mid = Math.floor((start + end) / 2);
    const node = new TreeNode(nums[mid]);
    node.left = buildBST(nums, start, mid - 1);
    node.right = buildBST(nums, mid + 1, end);
    return node;
  }
  return buildBST([3], 0, nums.length - 1);
}

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

Anums

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names like 'array' causes reference errors.

The array parameter and recursive calls use the name 'nums' consistently.

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