原题链接在这里：https://leetcode.com/problems/binary-tree-right-side-view/

题目：

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]. 

题解：

与Binary Tree Level Order Traversal相似，通过BFS一层一层扫树，这里是仅添加最右面的一个点，就是curCount = 1 的时候.

Note: 当curCount == 0 时, curCount = nextCount, 不要忘记把nextCount归0.

Time Complexity: O(n). n是tree的node总数.

Space: O(n), que的大小.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<Integer> rightSideView(TreeNode root) {
12         List<Integer> res = new ArrayList<Integer>();
13         if(root == null){
14             return res;
15         }
16         
17         LinkedList<TreeNode> que = new LinkedList<TreeNode>();
18         que.add(root);
19         int curCount = 1;
20         int nextCount = 0;
21         while(!que.isEmpty()){
22             TreeNode cur = que.poll();
23             curCount--;
24             
25             if(cur.left != null){
26                 que.add(cur.left);
27                 nextCount++;
28             }
29             if(cur.right != null){
30                 que.add(cur.right);
31                 nextCount++;
32             }
33             
34             if(curCount == 0){
35                 res.add(cur.val);
36                 curCount = nextCount;
37                 nextCount = 0;
38             }
39         }
40         
41         return res;
42     }
43 }