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

Recall & Review

beginner

What is the main goal when converting a sorted array to a balanced BST?

To create a binary search tree where the height difference between left and right subtrees of any node is minimal, ensuring efficient search operations.

beginner

Why do we choose the middle element of the sorted array as the root in this conversion?

Choosing the middle element ensures that the left and right subtrees have roughly equal numbers of nodes, helping keep the tree 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 the tree nodes.

beginner

In the recursive approach, what are the base cases when building the BST?

When the start index is greater than the end index, it means no elements are left to process, so return nullptr (no node).

beginner

How does the recursive function divide the array to build left and right subtrees?

It recursively calls itself on the left half (start to mid-1) for the left subtree and on the right half (mid+1 to end) for the right subtree.

What element is chosen as the root node when converting a sorted array to a balanced BST?

A The first element

B The middle element

C The last element

D A random element

The middle element is chosen to keep the tree balanced by evenly dividing the array.

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

A O(n)

B O(n log n)

C O(log n)

D O(n^2)

Each element is processed once, so the time complexity is O(n).

What happens when the start index is greater than the end index in the recursive function?

A Restart the recursion

B Create a new node

C Throw an error

D Return null (no node)

This is the base case indicating no elements left to process, so return null.

Which traversal method is naturally used to build the BST from the sorted array?

A Preorder traversal

B Postorder traversal

C Inorder traversal

D Level order traversal

The recursive approach picks the middle element first (root), then builds left and right subtrees, resembling preorder.

Why is the resulting BST balanced after conversion from a sorted array?

A Because the array is sorted

B Because the recursion stops early

C Because the middle element is chosen as root at each step

D Because the array size is even

Choosing the middle element ensures left and right subtrees have similar sizes, keeping the tree balanced.

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

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

Describe why the balanced BST created from a sorted array is efficient for search operations.

Consider how tree height affects search speed.