LeetCode 230: Kth Smallest Element in a BST

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?

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).

解题思路

思路一：返回中序遍历下第 k 个元素。下面代码利用了二叉树中序遍历的非递归算法：

/**
* 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) {
stack<TreeNode *> ns;
TreeNode *p = root;
int count = 0;
while (p != NULL || !ns.empty()) {
while (p != NULL) {
ns.push(p);
p = p->left;
}
if (!ns.empty()) {
p = ns.top();
ns.pop();

if (++count == k) return p->val;

p = p->right;
}
}
}
};

思路二：如果

TreeNode

有一个属性

leftCnt

记录了左子树的节点个数，则在查找时可以用二分查找，时间复杂度为 O(heightofBST)。

记当前节点为node

当node不为空时循环：

若k == node.leftCnt + 1：则返回node

否则，若k > node.leftCnt：则令k -= node.leftCnt + 1，令node = node.right

否则，node = node.left