【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?

這道題其實很好地利用BST的特點就好了，舉個栗子：

     2

   /     \

1         3

我怎麼找到第kth的？如果我能排序的話，是不是特別簡單。所以我一眼就看出來了，這其實不就是中序遍歷放進一個queue就好了麼。。。

當然，這裏要注意，queue和stack在java裏不一樣，stack可以直接new，而queue是一個interface，所以new的時候只能 new LinkedList()或者new ArrayDeque();

然後就是巧用queue做中序遍歷：

package testAndfun;

import java.util.ArrayDeque;
import java.util.Queue;


public class KthSmallestEle {
	Queue<Integer> queue = new ArrayDeque<Integer>();
    
	
    public int kthSmallest(TreeNode root, int k){
        foo(root);
        while(k>1){
            queue.poll();
            k--;
        }
        return queue.poll();
    }
    
    public void foo(TreeNode root) {
    	 if(root.left!=null){
             foo(root.left);
             queue.add(root.val);
             if(root.right!=null){
                 foo(root.right);
             }
         }
         else{
             queue.add(root.val);
             if(root.right!=null){
                 foo(root.right);
             }
         }
    }
    
    class TreeNode{
    	int val;
    	TreeNode left;
    	TreeNode right;
    	TreeNode(int x){
    		val = x;
    	}
    }
}