The Art of Balancing a Binary Search Tree (BST)

Transforming an unbalanced BST into a balanced one might seem like a daunting task, but it's simpler than you might think. This process gains particular importance when we want to optimize the performance of operations like search, insert, and delete. Specialized trees like Red-Black Trees or AVL trees balance themselves during these operations. But what if you're handed an unbalanced tree to start with?

The Significance of Using an Array

Arrays are a versatile data structure and can be incredibly efficient when it comes to sorting and retrieving data. In the context of balancing a BST, an array serves as a temporary holding place for the tree's elements, sorted in ascending order. Once we have this sorted array, creating a balanced tree becomes a straightforward task.

How We Arrive at This Strategy

Balancing a tree involves ensuring that the left and right subtrees of each node differ in height by at most one. A sorted array inherently satisfies this constraint when converted back into a BST, making it an excellent strategy for our purpose.

The Two-Step Algorithm: Insight and Intuition

Here are the two key steps, along with the intuition behind them:

Step 1: In-Order Traversal into an Array

• What We Do: Perform an in-order traversal of the given unbalanced BST and store each node's value in an array.
• Why We Do It: An in-order traversal of a BST naturally produces a sorted list of its elements. This array now contains all elements of the BST in their proper order but without the baggage of the tree's previous imbalances.

Step 2: Create a Balanced BST from the Sorted Array

• What We Do: Recursively pick the middle element from the sorted array to be the root and repeat the process for the sub-arrays to form left and right subtrees.
• Why We Do It: Choosing the middle element ensures that the tree remains balanced. The left and right halves of the array naturally form the left and right subtrees of each node, maintaining the balanced property.