关于修剪二叉搜索树中的递归思路

/**
 * 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==nullptr){
            return root;
        }

        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;
    }
};

这里在面对不符合条件的节点时，采取的方式是跳过当前节点然后处理后面的节点。然后底下面对符合条件的节点也要对它的左子树和右子树进行遍历