Given a Binary Search tree and an integer K, we have to split the tree into two Balanced Binary Search Tree, where BST-1 consists of all the nodes which are less than K and BST-2 consists of all the nodes which are greater than or equal to K.
Note: Arrangement of the nodes may be anything but both BST should be Balanced.
Input: 40 / \ 20 50 / \ \ 10 35 60 / / 25 55 K = 35 Output: First BST: 10 20 25 Second BST: 35 40 50 55 60 Explanation: After splitting above BST about given value K = 35 First Balanced Binary Search Tree is 20 / \ 10 25 Second Balanced Binary Search Tree is 50 / \ 35 55 \ \ 40 60 OR 40 / \ 35 55 / \ 50 60 Input: 100 / \ 20 500 / \ 10 30 \ 40 K = 50 Output: First BST: 10 20 30 40 Second BST: 100 500 Explanation: After splitting above BST about given value K = 50 First Balanced Binary Search Tree is 20 / \ 10 30 \ 40 Second Balanced Binary Search Tree is 100 \ 500
- First store the inorder traversal of given BST in an array
- Then, split this array about given value K
- Now construct first balanced BST by first splitting part and second BST by second splitting part, using the approach used in this article.
Below is the implementation of the above approach:
Original BST : 2 3 4 5 6 7 8 First BST : 2 3 4 Second BST : 5 6 7 8
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- Check for Identical BSTs without building the trees
- Construct all possible BSTs for keys 1 to N
- Find pairs with given sum such that pair elements lie in different BSTs
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