题目：
Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.
Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.
Example 1:
Input: root = [1,0,2], low = 1, high = 2 Output: [1,null,2]
Example 2:
Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3 Output: [3,2,null,1]
Example 3:
Input: root = [1], low = 1, high = 2 Output: [1]
Example 4:
Input: root = [1,null,2], low = 1, high = 3 Output: [1,null,2]
Example 5:
Input: root = [1,null,2], low = 2, high = 4 Output: [2]
Constraints:
- The number of nodes in the tree in the range
[1, 10^4]. 0 <= Node.val <= 10^4- The value of each node in the tree is unique.
-
rootis guaranteed to be a valid binary search tree. 0 <= low <= high <= 10^4
思路：
有两种方法，一种比较繁琐的是先读一遍整棵树，找到在[low, high]中的对应点并记录下来，可以用哈希表也可以用vector，然后再重建一个树即可。这里介绍思路更简单的递归，首先如果当前node为空直接返回，之后因为我们最后要返回root，所以最后处理左和右，再返回本身的root。接下来分类讨论：两种情况。如果当前值在low和high之外：如果当前值小于low，那么root的左子肯定更小，不符合，我们只要递归root的右子即可；如果当前值大于high，相应的，我们只要递归root的左子即可。如果当前值在low和high之内，直接处理左右子即可。最后返回root本身。
代码：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int low, int high) {
if (!root) return NULL;
if (root->val < low)
return trimBST(root->right, low, high);
if (root->val > high)
return trimBST(root->left, low, high);
root->left = trimBST(root->left, low, high);
root->right = trimBST(root->right, low, high);
return root;
}
};