[LeetCode]Split BST

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.

题目大意:

将二叉排序树拆成两棵子树,左子树的所有节点均不大于V,右子树的所有节点均大于V

解题思路:

递归(Recursion)

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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