Split BST

Intuition

We could solve this problem recursivly. For each tree, we could always get the sub-tree who are greater than the key and another sub-tree who is smaller or equal to the original tree.

Solution


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def splitBST(self, root: TreeNode, V: int) -> List[TreeNode]:
        if root is None:
            return [None, None]
        if root.val > V:
            l = self.splitBST(root.left, V)
            root.left = l[1]
            return [l[0], root]
        else:
            r = self.splitBST(root.right, V)
            root.right = r[0]
            return [root, r[1]]