230. Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

Input: root = [3,1,4,null,2], k = 1
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Constraints:

• The number of nodes in the tree is n.
• 1 <= k <= n <= 104
• 0 <= Node.val <= 104

题目：给定一个二叉查找树，返回第K小的值

思路：树的inorder遍历的变体。

方法一：迭代法，代码：

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        int count = 0;
        stack stk;
        TreeNode* node = root;
        while(!stk.empty() || node){
            while(node){
                stk.push(node);
                node = node->left;
            }
            node = stk.top();
            stk.pop();
            count++;
            if(count == k) return node->val;
            node = node->right;
        }
        return -1;
    }
};

方法二，递归法，代码：

/**
 * 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:
    bool DFS(TreeNode* node, int k, int& pos, int& res){
        if(!node || pos > k) return false;
        bool left = DFS(node->left, k, pos, res);
        if(left) return true;
        pos++;
        if(pos == k){
            res = node->val;
            return true;
        }
        return DFS(node->right, k, pos, res);
        
    }
    int kthSmallest(TreeNode* root, int k) {
        int pos = 0, res = 0;
        DFS(root, k, pos, res);
        return res;
    }
};