Convert Sorted Array to Balanced BST in DSA Go - 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 pick the middle element of the sorted array as the root in this conversion?

Choosing the middle element as root divides the array into two halves of roughly equal size, helping keep the tree balanced.

intermediate

In the recursive approach to build a balanced BST from a sorted array, what are the subproblems?

Building the left subtree from the left half of the array and the right subtree from the right half of the array.

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 nodes.

beginner

What is the base case in the recursive function to convert a sorted array to a balanced BST?

When the start index is greater than the end index, meaning there are no elements to process, so return nil.

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

A The middle element

B The first element

C The last element

D A random element

The middle element is chosen to keep the tree balanced by dividing the array into two equal halves.

What is the main advantage of a balanced BST over an unbalanced BST?

A Easier to implement

B Uses less memory

C Faster search, insert, and delete operations

D Allows duplicate values

Balanced BSTs keep operations like search, insert, and delete efficient by maintaining minimal height.

Which traversal method is naturally used when converting a sorted array to a balanced BST?

A Preorder traversal

B Level order traversal

C Postorder traversal

D Inorder traversal

The sorted array corresponds to the inorder traversal of the BST, so picking the middle element simulates inorder construction.

What happens if you always pick the first element as root when converting a sorted array to BST?

A The BST becomes skewed and unbalanced

B The BST is perfectly balanced

C The BST has random shape

D The BST is empty

Picking the first element repeatedly creates a skewed tree similar to a linked list, losing balance.

What is the stopping condition for the recursive function building the BST from a sorted array?

A When start index == end index

B When start index > end index

C When array is empty

D When root is null

The recursion stops when the start index exceeds the end index, meaning no elements remain to process.

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

Think about dividing the array and building subtrees.

Why is it important to keep the BST balanced when converting from a sorted array? What problems arise if it is not balanced?

Consider how tree height affects operation speed.