Convert Sorted Array to Balanced BST in DSA Javascript - 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 ensures that the left and right subtrees have roughly equal numbers of nodes, keeping 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 nodes.

beginner

In the recursive approach, what are the base cases when converting a sorted array to a balanced BST?

When the start index is greater than the end index, it means there are no elements to process, so return null.

beginner

How does the recursive function build the left and right subtrees?

It recursively calls itself on the left half of the array for the left subtree and on the right half for the right subtree.

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

A The first element

B The last element

C The middle element

D A random element

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

What is the base case in the recursive function for this conversion?

A When the middle element is null

B When start index equals end index

C When the array is empty

D When start index is greater than end index

The recursion stops when the start index is greater than the end index, meaning no elements remain.

What is the time complexity of converting a sorted array to a balanced BST?

A O(n log n)

B O(n)

C O(n^2)

D O(log n)

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

Which traversal order is naturally followed when building the BST from the sorted array?

A Preorder

B Inorder

C Postorder

D Level order

The root is created first, then left and right subtrees, matching preorder traversal.

Why is the resulting BST balanced?

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

B Because duplicates are removed

C Because the tree is built from left to right

D Because the array is sorted

Choosing the middle element as root ensures left and right subtrees are 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 balanced trees keep search times low.