#leetcode#Kth Smallest Element in a BST，

#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?

思路， in order traversal，

注意第一個元素的判斷， 設root為當前節點， 則 root.left == null && count == 0，就保證了root為 1st smallest element，例子：

   4

  /

/

     2

  /

/

     1

    4

  /

/

       2

         \

           \

    3

上面兩個bst中第一個節點分別為 1 和 2

/** * 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) {        List<Integer> count = new ArrayList<Integer>();        count.add(0);        helper(root, count, k);        return count.get(1);    }        private void helper(TreeNode root, List<Integer> count, int target){        if(root == null){            return;        }                helper(root.left, count, target);                if(root.left == null && count.get(0) == 0){            count.set(0, 1);        }else{            count.set(0, count.get(0) + 1);        }        if(count.get(0) == target){            count.add(root.val);            return;        }                helper(root.right, count, target);    }}

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