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?
Hint:
Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
如果树节点结构是这样的话,直接进行中序遍历得出,如果在树中的节点添加一个count字段用于表示该子树含有的数值数量就可以在logn的时间里得出。就是一个动态计算rank的功能。
下面是inorder的代码:
/**
* 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 {
private:
int idx;
int value;
int K;
public:
int kthSmallest(TreeNode* root, int k) {
K = k;
idx = 0;
dfs(root);
return value;
}
void dfs(TreeNode* root) {
if (root == NULL) {
return;
}
dfs(root->left);
idx++;
if (idx == K) {
value = root->val;
return;
}
dfs(root->right);
}
};