669. 修剪二叉搜索树

根据根节点的值和区间进行对比，来判断递归哪个子树，

  1. class Solution {
  2. public:
  3.     TreeNode* trimBST(TreeNode* root, int low, int high) {
  4.         if(root==NULL)return root;
  5.         if(root->val < low)return trimBST(root->right,low,high);
  6.         else if(root->val >high)return trimBST(root->left,low,high);
  7.         else{
  8.             if(root->left)root->left = trimBST(root->left,low,high);
  9.             if(root->right)root->right = trimBST(root->right,low,high);
  10.         }
  11.         return root;
  12.     }
  13. };