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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

mid := (start + [1]) / 2

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Aend

Blen(arr)

C0

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(arr) instead of end index.

Using start twice instead of end.

The middle index is calculated by averaging the start and end indices.

2fill in blank

medium

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

DSA Go

node := &TreeNode{Val: [1]

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Aarr[mid]

Barr[end]

Carr[start]

Darr[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr[start] or arr[end] instead of arr[mid].

Using arr[0] which is always the first element.

The root node value is the middle element of the current array segment.

3fill in blank

hard

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

DSA Go

node.Left = sortedArrayToBST(arr, start, [1])

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Astart + 1

Bmid

Cmid - 1

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid as end index includes the middle element again.

Using end includes elements after mid.

The left subtree uses the array segment before the middle element, so end index is mid - 1.

4fill in blank

hard

Fill both blanks to build the right subtree correctly.

DSA Go

node.Right = sortedArrayToBST(arr, [1], [2])

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Amid + 1

Bstart

Cend

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid as start includes middle element again.

Using start or mid incorrectly as end index.

The right subtree uses the array segment after the middle element, from mid + 1 to end.

5fill in blank

hard

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

DSA Go

if [1] > [2] {
    return nil
}

mid := (start + end) / 2
node := &TreeNode{Val: arr[mid]}
node.Left = sortedArrayToBST(arr, start, [3])
return node

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

Astart

Bend

Cmid - 1

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using start >= end in base case instead of >.

Using mid or mid + 1 for left subtree end index.

The base case checks if start > end to stop recursion. The left subtree uses end index mid - 1.