230 Kth Smallest Element in a BST

题目链接：https://leetcode.com/problems/kth-smallest-element-in-a-bst/

题目：

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：
这题的考点是二分查找树的中序遍历。
一提到 BST ，首先要想到其中序遍历正好是数字的从小到大的排序。
中序遍历可以用递归，也可以使用栈。

代码中使用栈来实现中序遍历。

代码实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        LinkedList stack = new LinkedList();
        int result = 0;

        while(!stack.isEmpty() || root != null) {
            if(root != null) {
                stack.push(root);
                root = root.left;
            } else {
                root = stack.pop();
                k --;
                if(k == 0)
                    result = root.val;
                root = root.right;
            }
        }
        return result;
    }
}

91 / 91 test cases passed.
Status: Accepted
Runtime: 7 ms