[Google Question][LeetCode] Split BST

Problem: Given a Binary Search Tree (BST) with root node root, and a target value V, split the tree into two subtrees where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all nodes that are greater than the target value.  It's not necessarily the case that the tree contains a node with value V.

Additionally, most of the structure of the original tree should remain.  Formally, for any child C with parent P in the original tree, if they are both in the same subtree after the split, then node C should still have the parent P.

You should output the root TreeNode of both subtrees after splitting, in any order.

Example:

Input: root = [4,2,6,1,3,5,7], V = 2
Output: [[2,1],[4,3,6,null,null,5,7]]
Explanation:
Note that root, output[0], and output[1] are TreeNode objects, not arrays.

The given tree [4,2,6,1,3,5,7] is represented by the following diagram:

          4
        /   \
      2      6
     / \    / \
    1   3  5   7

while the diagrams for the outputs are:

          4
        /   \
      3      6      and    2
            / \           /
           5   7         1

Note:

• The size of the BST will not exceed 50.
• The BST is always valid and each node's value is different.

Approach: Let's take a hint from the question description. At any node there are two conditions:

1. curr_node.Value <= V: In that case we know that the whole left subtree of curr_node including the curr_node is going to be part of first half but there is possibility that there are few nodes in the right subtree of curr_node which belongs to first half and other nodes of right subtree belongs to the second half. That means we need to make a recursive call for the right subtree,
2. curr_node.Value > V: In this case we are sure that right subtree of curr_node including the curr_node is part of second half but we are not sure about left subtree yet. That means we need to make recursive call for left subtree.

If you understand the above two points the implementation is simple. Just have a look!

Implementation in C#:

    public TreeNode[] SplitBST(TreeNode root, int V) 

    {

        if (root == null)

        {

            return new TreeNode[] { null, null };

        }    

        if (root.val <= V)

        {

            TreeNode[] bsts = this.SplitBST(root.right, V);

            root.right = bsts[0];

            bsts[0] = root;

            return bsts;

        }

        else

        {

            TreeNode[] bsts = this.SplitBST(root.left, V);

            root.left = bsts[1];

            bsts[1] = root;

            return bsts;

        }

    }

Complexity: O(height of BST)