【LeetCode】230. Kth Smallest Element in a BST (2 solutions)
Kth Smallest Element in a BST
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.
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?
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
解法一:递归中序遍历,必须全部遍历
/**
* 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) {
vector<int> ret;
inOrder(root, ret);
return ret[k-];
}
void inOrder(TreeNode* root, vector<int>& ret)
{
if(root)
{
inOrder(root->left, ret);
ret.push_back(root->val);
inOrder(root->right, ret);
}
}
};
解法二:迭代中序遍历,遍历到第k个元素停止
/**
* 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) {
vector<int> ret;
stack<TreeNode*> stk;
stk.push(root);
TreeNode* cur = root;
while(cur->left)
{
stk.push(cur->left);
cur = cur->left;
}
while(!stk.empty())
{
TreeNode* top = stk.top();
stk.pop();
ret.push_back(top->val);
if(ret.size() == k)
break;
if(top->right)
{
TreeNode* cur = top->right;
stk.push(cur);
while(cur->left)
{
stk.push(cur->left);
cur = cur->left;
}
}
}
return ret[k-];
}
};
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