leetcode算法：Trim a Binar Search Tree

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1:
Input: 
    1
   / \
  0   2

  L = 1
  R = 2

Output: 
    1
      \
       2
Example 2:
Input: 
    3
   / \
  0   4
   \
    2
   /
  1

  L = 1
  R = 3

Output: 
      3
     / 
   2   
  /
 1

这道题描述的需求是：
给我们一个二叉排序树 最小数l 和最大数r
我们需要做的是对这棵树进行裁剪，让树里所有的节点值都满足在l和r之间

思想：
二叉排序树的特点是：对于任意一个节点，左子树任意节点数值都比跟节点小，右子树任意节点数值都比根大
所以考虑，对于任意一个节点，
如果值比l还小，那就应该抛弃这个根，去右子树寻找新的根
如果值比r还大，那就应该抛弃这个根，去左子树寻找新的根

这样的操作进行递归就可以了

我的python代码：

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

 class Solution:
     def trimBST(self, root, L, R):
         """
         :type root: TreeNode
         :type L: int
         :type R: int
         :rtype: TreeNode
         """
         if root is None:
             return None
         if root.val >R:
             return self.trimBST(root.left ,L,R)
         if root.val<L:
             return self.trimBST(root.right, L, R)
         root.left = self.trimBST(root.left,L,R)
         root.right = self.trimBST(root.right,L,R)
         return root

 if __name__ == '__main__':
     s = Solution()

     root = TreeNode(1)
     root.left = TreeNode(0)
     root.right = TreeNode(2)

     print(root.val,root.left.val,root.right.val)

     root = s.trimBST(root,1,2)

     print(root, root.left, root.right)