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.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3

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1 : 最小堆

Nth Smallest/ Top Kth Element都屬于用PriorityQueue可以解決問題的范疇。

1. 遍歷樹，用最小堆來存K個node。
2. 最小堆頂?shù)哪莻€node就是Nth Smallest

思路2： in-order traversal

對于BST，in-order遍歷時，第K個訪問的節(jié)點就是結果。

1. PriorityQueue Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        if (root == null) return 0;
        
        Comparator<TreeNode> cmp = new Comparator<TreeNode>() {
            public int compare(TreeNode i1, TreeNode i2) {
                return i2.val - i1.val;
            }
        };
        
        PriorityQueue<TreeNode> queue = new PriorityQueue<TreeNode>(k, cmp);
        traverseTree(root, queue, k);
        
        return queue.peek().val;
    }
    
    private void traverseTree(TreeNode root, PriorityQueue<TreeNode> priorityQueue, int k) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        
        while (!queue.isEmpty()) {
            TreeNode curNode = queue.poll();
            priorityQueue.add(curNode);
            if (priorityQueue.size() > k) {
                priorityQueue.poll();
            }
            
            if (curNode.left != null) {
                queue.add(curNode.left);
            }
            
            if (curNode.right != null) {
                queue.add(curNode.right);
            }
        }
    }
}

2. In-Order Traversal Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        if (root == null) {
            return 0;
        }
        
        int[] counter = { k };
        
        return getTarget (root, counter).val;
    }
    
    private TreeNode getTarget (TreeNode root, int[] counter) {
        TreeNode target = null;
        
        if (root.left != null) {
            target = getTarget (root.left, counter);
        }
        
        if (target == null) {
            if (counter[0] == 1) {
                target = root;
            }
            
            counter[0] --;
        }
        
        if (target == null && root.right != null) {
            target = getTarget (root.right, counter);
        }
        
        return target;
    }
}