## 题目描述：

LeetCode 776. Split BST

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 1:

```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:

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

## Python代码：

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

class Solution(object):
def splitBST(self, root, V):
"""
:type root: TreeNode
:type V: int
:rtype: List[TreeNode]
"""
if not root: return [None, None]
if root.val > V:
left, right = self.splitBST(root.left, V)
root.left = right
return [left, root]
left, right = self.splitBST(root.right, V)
root.right = left
return [root, right]
``````

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