题目描述
给定一个二叉搜索树的根节点 root ,和一个整数 k ,请你设计一个算法查找其中第 k 小的元素(k 从 1 开始计数)。
示例 1:

输入: root = [3,1,4,null,2], k = 1
输出: 1
示例 2:
输入: root = [5,3,6,2,4,null,null,1], k = 3
输出: 3
提示:
树中的节点数为 n 。
1<=k<=n<=1041 <= k <= n <= 10^41<=k<=n<=104
0<=Node.val<=1040 <= Node.val <= 10^40<=Node.val<=104
思路
中序遍历前k个。
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 res = 0; // 记录答案
TreeNode *predecessor; // root的前一个结点
while(root != nullptr)
{
if(root->left != nullptr)
{
predecessor = root->left;
while(predecessor->right != nullptr && predecessor->right != root)
{
predecessor = predecessor->right;
}
// 第一次标记root
if(predecessor->right == nullptr)
{
predecessor->right = root;
root = root->left;
}
// 第二次遍历结束
else
{
predecessor->right = nullptr;
—k;
if(k == 0) res = root->val;
root = root->right;
}
}
else
{
—k;
if(k == 0) res = root->val;
root = root->right;
}
}
return res;
}
};


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