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 + (end - start) / [1];

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A1

B3

C4

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using division by 3 or 4 instead of 2.

Forgetting to subtract start from end before dividing.

The middle index is found by dividing the length by 2 to split the array into two halves.

2fill in blank

medium

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

DSA C++

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

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Alist

Bvector

Carr

Dtree

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable name like 'list' or 'vector'.

Using 'tree' which is unrelated.

The array is named 'arr', so we access the middle element using arr[mid].

3fill in blank

hard

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

DSA C++

node->left = buildBST(arr, start, [1] - 1);

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Amid

Bstart

Cend

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using start or end incorrectly as the end index.

Using mid + 1 which belongs to the right subtree.

The left subtree includes elements before the middle, so end index is mid - 1.

4fill in blank

hard

Fill both blanks to build the right subtree correctly.

DSA C++

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

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Amid + 1

Bstart

Cend

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using start or mid - 1 as start index for right subtree.

Swapping start and end indices.

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

5fill in blank

hard

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

DSA C++

if (start > [1]) return nullptr;
int mid = start + (end - start) / [2];
TreeNode* node = new TreeNode(arr[mid]);
node->left = buildBST(arr, start, mid - 1);
node->right = buildBST(arr, mid + 1, [3]);
return node;

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Aend

B2

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using start instead of end in base case.

Dividing by wrong number in mid calculation.

Using wrong end index for right subtree.

Base case checks if start is greater than end. Mid is calculated by dividing by 2. Right subtree ends at end.