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]

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

這道題要求我們打印出二叉樹每一行最右邊的一個數字，實際上是求二叉樹層序遍歷的一種變形，我們只需要儲存每一層最右邊的數字即可，可以參考我之前的部落格 Binary Tree Level Order Traversal 二叉樹層序遍歷，這道題只要在之前那道題上稍加修改即可得到結果，還是需要用到資料結構佇列queue，遍歷每層的節點時，把下一層的節點都存入到queue中，每當開始新一層節點的遍歷之前，先把新一層最後一個節點值存到結果中，程式碼如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int
 
> res;
        if (!root) return res;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            res.push_back(q.back()->val);
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }
};