Complexity of constructing a balanced BST from a given normal BST

To understand the complexity of constructing a balanced Binary Search Tree (BST) from a given normal BST, let's break down the process and analyze each step.

1. Inorder Traversal:
The first step is to perform an inorder traversal of the given BST to obtain a sorted array of the elements. Inorder traversal visits the nodes in ascending order, which ensures that the sorted array obtained will be in non-decreasing order.

The time complexity of performing an inorder traversal on a BST is O(n), where n is the number of nodes in the BST. This is because we need to visit each node exactly once.

2. Constructing the Balanced BST:
Once we have the sorted array, we can construct a balanced BST by recursively dividing the array into two halves and setting the middle element as the root of the subtree. This process is similar to constructing a binary search tree using the divide and conquer approach.

The time complexity of constructing a balanced BST from a sorted array is O(n), where n is the number of elements in the array. This is because we need to visit each element in the array once to construct the BST.

Overall Complexity:
Considering both steps, the time complexity of constructing a balanced BST from a given normal BST is O(n), where n is the number of nodes in the original BST.

Explanation:
The complexity is O(n) because we only need to perform an inorder traversal of the BST and construct a balanced BST from the sorted array. Both steps have a time complexity of O(n), so the overall complexity is also O(n).

Conclusion:
The efficient algorithm to construct a balanced BST from a given normal BST has a time complexity of O(n), where n is the number of nodes in the original BST. This algorithm ensures that the resulting BST is balanced, which improves the efficiency of various operations performed on the tree.