思路：二叉搜索树的中序遍历是有序的从大到小的，故得出中序遍历的结果，即要第cnt大的数为倒数第cnt的数

/**
 * 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:
    //用于保存中序遍历结果
    vector<int> inorder;
    int findTargetNode(TreeNode* root, int cnt) {
        //进行中序遍历
        getorder(root);
        //返回中序遍历中倒数第cnt个数即为所求的结果
        return inorder[inorder.size()-cnt];
    }
    //进行中序遍历
    void getorder(TreeNode* root){ 
        if(root == nullptr) return;
        getorder(root->left);
        inorder.push_back(root->val);
        getorder(root->right);
    }
};