题目：

题解：

class MyBst {
public:
    MyBst(TreeNode *root) {
        this->root = root;
        countNodeNum(root);
    }

    // 返回二叉搜索树中第k小的元素
    int kthSmallest(int k) {
        TreeNode *node = root;
        while (node != nullptr) {
            int left = getNodeNum(node->left);
            if (left < k - 1) {
                node = node->right;
                k -= left + 1;
            } else if (left == k - 1) {
                break;
            } else {
                node = node->left;
            }
        }
        return node->val;
    }

private:
    TreeNode *root;
    unordered_map<TreeNode *, int> nodeNum;

    // 统计以node为根结点的子树的结点数
    int countNodeNum(TreeNode * node) {
        if (node == nullptr) {
            return 0;
        }
        nodeNum[node] = 1 + countNodeNum(node->left) + countNodeNum(node->right);
        return nodeNum[node];
    }

    // 获取以node为根结点的子树的结点数
    int getNodeNum(TreeNode * node) {
        if (node != nullptr && nodeNum.count(node)) {
            return nodeNum[node];
        }else{
            return 0;
        }
    }
};

class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        MyBst bst(root);
        return bst.kthSmallest(k);
    }
};