LeetCode OJ：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].

思路：

右子树优先遍历，同时计算遍历过的层次，遍历完右子树后开始遍历左子树，若左子树的深度比右子树的深，则把左子树更深层次的节点也加入结果集里面

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void rightSideView(TreeNode *root, vector<int> &res, int n){
        if(!root) return;
        ++n;
        if(n == res.size() + 1){
            res.push_back(root->val);
        }
        rightSideView(root->right, res, n);
        rightSideView(root->left, res, n);
    }
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int> res;
        rightSideView(root, res, 0);
        return res;
    }
};