难度中等344
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
个最小元素(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:
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- 树中的节点数为
n
。1 <= k <= n <= 104
0 <= Node.val <= 104
进阶:如果二叉搜索树经常被修改(插入/删除操作)并且你需要频繁地查找第
k
小的值,你将如何优化算法?
1、递归
/**
* 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) {
vector<int> vec;
midTraverse(root, vec);
return vec[k-1];
}
void midTraverse(TreeNode* root, vector<int>& vec){
if(root == nullptr){
return;
}
midTraverse(root->left, vec);
vec.push_back(root->val);
midTraverse(root->right, vec);
}
};
2、迭代
/**
* 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) {
stack<TreeNode*> stk;
while(root != nullptr || !stk.empty()){
while(root != nullptr){
stk.push(root);
root = root->left;
}
root = stk.top();
stk.pop();
k --;
if(k == 0){
return root->val;
}
root = root->right;
}
return 0;
}
};