Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

思路分析：这题本质上是考察树的遍历，读完题目要能准备理解题意就是要返回每一个level最右边的结点，所以想到用BFS，借助队列实现，但是每一层从右向左扫描，同时记住index，每一个level index为0的node就是需要保存到结果之中的node。时间复杂度O（N），空间复杂度也是O（N），因为需要额外队列保存结点。

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 List rightSideView(TreeNode root) {

    List res = new ArrayList();
    if(root == null) {
        return res;
    }
    
    LinkedList queue = new LinkedList();
    queue.add(root);
    while(!queue.isEmpty()){
        int numNodeInCurLevel = queue.size();
        for(int i = 0; i < numNodeInCurLevel; i++){
            TreeNode node = queue.poll();
             if(i == 0) res.add(node.val);
            if(node.right != null){
                queue.add(node.right);
            }
            if(node.left != null){
                queue.add(node.left);
            }
        }
    }
    return res;   
} 
}