LeetCode 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:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

思路分析:基本就是实现BST的中序遍历。BST中序遍历得到有序的数组，能够easy知道第k小数。下面给出了递归实现，借助counter记录当前遍历过的node数目。当counter==k时就能够返回。

当然也能够借助栈实现迭代的解法。

Follow up： 进一步优化。我们能够在节点中额外保留一些信息: 左子树的大小. 在插入删除时也同一时候维护左子树的大小.进行查找时能够比較左子树size和k的大小，就能够知道要找的node在左边还是右边。通过分治法的思路加速. 时间复杂度为O(h)。

下面的伪代码摘录自 http://bookshadow.com/weblog/2015/07/02/leetcode-kth-smallest-element-bst/

假设BST节点TreeNode的属性能够扩展，则再加入一个属性leftCnt，记录左子树的节点个数

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
上述算法时间复杂度为O(BST的高度)

AC Code

/**
 * 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 res = 0;
    public int counter = 0;
    
    public int kthSmallest(TreeNode root, int k) {
        inOrderK(root, k);
        return res;
    }
    
    public void inOrderK(TreeNode root, int k){
        if(root.left!=null) inOrderK(root.left, k);
        counter++;
        
        if(counter == k) {
            res = root.val;
            return;
        }
        if(root.right!=null) inOrderK(root.right, k);
    }
}