*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?

思路：

就是inorder traverse

 1     public int kthSmallest(TreeNode root, int k) 
 2     {
 3         
 4         ArrayList<Integer> re = new ArrayList<Integer>();
 5        // if(root==null)
 6     //        return re;
 7         helper(root,re);
 8         return re.get(k-1);
 9 
10     }
11     
12     public void helper(TreeNode root, ArrayList<Integer> re)
13     {
14         if(root==null)
15             return;
16         helper(root.left,re);
17         re.add(root.val);
18         helper(root.right,re);
19     }

reference：http://www.programcreek.com/2014/07/leetcode-kth-smallest-element-in-a-bst-java/

解法二：非递归

/**
 * 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) 
    {
        ArrayList<Integer> res = inOrderTrav(root);
        return res.get(k-1);
        
    }
    
    public ArrayList<Integer> inOrderTrav(TreeNode root)
    {
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root==null) return res;
        while(root!=null||!stack.isEmpty())
        {
            if(root!=null)
            {
                stack.push(root);
                root = root.left;
            }
            else
            {
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}