题目链接
tag:
- Medium；
- Stack；
question：
Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
Example 1:
Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3 6
/ \
2 4
/
1
Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
思路：
这道题给的提示是让我们用BST的性质来解题，最重要的性质是就是左<根<右，那么如果用中序遍历所有的节点就会得到一个有序数组。所以解题的关键还是中序遍历啊。关于二叉树的中序遍历可以参见我之前的博客Binary Tree Inorder Traversal 二叉树中序遍历，我们来看一种非递归的方法，中序遍历最先遍历到的是最小的结点，那么我们只要用一个计数器，每遍历一个结点，计数器自增1，当计数器到达k时，返回当前结点值即可，代码如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
int cnt = 0;
stack<TreeNode*> sk;
TreeNode* p = root;
while (p || !sk.empty()) {
while (p) {
sk.push(p);
p = p->left;
}
p = sk.top();
sk.pop();
++cnt;
if (cnt == k)
return p->val;
p = p->right;
}
return 0;
}
};
