LeetCode 669. Trim a Binary Search Tree

题目：
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   2L = 1R = 2

Output:

1\2

Example 2:
Input:

    3/ \
  0   4\
    2/1L = 1R = 3

Output:

      3/ 2   /1

思路：如果节点为空，直接返回NULL；如果节点不为空，如果当前节点的值小于L，递归右子节点；如果当前节点的值大于R，递归左子节点；如果介于L、R之间，那么root的左右子节点分别递归。

代码：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:TreeNode* trimBST(TreeNode* root, int L, int R) {if(root){//如果节点不为空if(root->val<L){//如果当前节点的值小于L，递归右子节点return trimBST(root->right,L,R);}if(root->val>R){//如果当前节点的值大于R，递归左子节点return trimBST(root->left,L,R);}root->left=trimBST(root->left,L,R);//如果介于L、R之间，那么root的左右子节点分别递归root->right=trimBST(root->right,L,R);return root;}else{//如果节点为空，直接返回NULLreturn NULL;}}
};

结果： 18ms

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