Convert a normal BST to Balanced BST
Given a BST (Binary Search Tree) that may be unbalanced, convert it into a balanced BST that has minimum possible height.
Input: 30 / 20 / 10 Output: 20 / \ 10 30 Input: 4 / 3 / 2 / 1 Output: 3 3 2 / \ / \ / \ 1 4 OR 2 4 OR 1 3 OR .. \ / \ 2 1 4 Input: 4 / \ 3 5 / \ 2 6 / \ 1 7 Output: 4 / \ 2 6 / \ / \ 1 3 5 7
A Simple Solution is to traverse nodes in Inorder and one by one insert into a self-balancing BST like AVL tree. Time complexity of this solution is O(n Log n) and this solution doesn’t guarantee
An Efficient Solution can construct balanced BST in O(n) time with minimum possible height. Below are steps.
- Traverse given BST in inorder and store result in an array. This step takes O(n) time. Note that this array would be sorted as inorder traversal of BST always produces sorted sequence.
- Build a balanced BST from the above created sorted array using the recursive approach discussed here. This step also takes O(n) time as we traverse every element exactly once and processing an element takes O(1) time.
Below is the implementation of above steps.
Preorder traversal of balanced BST is : 7 5 6 8 10
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