Given a binary tree, imagine yourself standing on therightside 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]

題目大意：給出一棵二叉樹，假設你現在站在它的右側，輸出你看到的節點（即二叉樹的右檢視）。

解題思路：二叉樹層序遍歷，輸出每一層的最後一個節點即可。因為層序遍歷有遞迴和非遞迴兩種寫法，所以本題也給出了兩種解法。

程式碼如下：

/**
 * Definition for a binary tree node.
 * 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) {
        // levelOrderTraversal(root, ans);
        _levelOrderTraversal(root, 0, ans);
        return ans;
    }
private:
    queue<TreeNode*> que;
    vector<int> ans;
    //非遞迴版本
    void levelOrderTraversal(TreeNode* root, vector<int>& ans) {
        if(root == nullptr) return ;
        int cur = 0, last, val;
        TreeNode* tmp;
        que.push(root);
        
        while(cur < que.size()) {
            last = que.size();
            while(cur < last) {
                tmp = que.front();
                que.pop();
                if(tmp->left) que.push(tmp->left);
                if(tmp->right) que.push(tmp->right);
                cur++;
            }
            ans.push_back(tmp->val);
            cur = 0;
        }
    }
    //遞迴版本
    void _levelOrderTraversal(TreeNode* root, int depth, vector<int>& ans) {
        if(root == nullptr) return ;
        if(depth >= ans.size()) ans.push_back(root->val);
        _levelOrderTraversal(root->right, depth + 1, ans);
        _levelOrderTraversal(root->left, depth + 1, ans);
    }
};