題目：

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].

解題思路：
這題的考點是樹的層次遍歷的變形。
1. 從右往左看去，能看到的每層的結點就是該層最右的結點，該層其它結點都被最右結點擋住了。
2. 解法就是層次遍歷每一層，每一層的最右結點就是要找的結點。
3. 即，使用一個佇列儲存樹的一層結點，將每一層最右結點放到結果連結串列中。

具體程式碼實現，採用兩個佇列，一個佇列放上一層的結點，一個佇列放當前層的結點。
也可以只採用一個佇列，此時只用記住每一層的結點個數，即可將一個佇列中的結點切分為父結點層和子結點層。

程式碼實現：
採用兩個佇列：

/**
 * 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<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList();
        if(root == null)
            return res;
        LinkedList<TreeNode> father = new LinkedList();
        father.add(root);
        res.add(root.val);
        while
 
(!father.isEmpty()) {
            LinkedList<TreeNode> children = new LinkedList();
            while(!father.isEmpty()) {
                TreeNode r = father.poll();
                if(r.left != null)
                    children.add(r.left);
                if(r.right != null)
                    children.add(r.right);
            }
            if(!children.isEmpty()) {
                father = children;
                res.add(father.peekLast().val);
            }
        }
        return res;
    }
}

210 / 210 test cases passed.
Status: Accepted
Runtime: 3 ms

只採用一個佇列：

/**
 * 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<Integer> rightSideView(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();

        if(root == null) return result;

        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);

        while(queue.size() > 0){
            //get size here
            int size = queue.size();

            for(int i=0; i<size; i++){
                TreeNode top = queue.remove();

                //the first element in the queue (right-most of the tree)
                if(i==0){
                    result.add(top.val);
                }
                //add right first
                if(top.right != null){
                    queue.add(top.right);
                }
                //add left
                if(top.left != null){
                    queue.add(top.left);
                }
            }
        }

        return result;
    }
}

210 / 210 test cases passed.
Status: Accepted
Runtime: 3 ms