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);
}
};