Binary Tree Right Side View

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：dfs。每次优先遍历右子树。同时需要记录当前所处的深度，当第一次进入一个深度时，就将该值存入结果。因为我们每次都是优先遍历右子树，因此这样做一定是最右侧能看到的节点。

class Solution {
public:
    void help(vector<int>& res, TreeNode* root, int depth)
    {
        if (!root) return;
        if (res.size() < depth + 1)
            res.push_back(root->val);
        help(res, root->right, depth + 1);
        help(res, root->left, depth + 1);
    }
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        help(res, root, 0);
        return res;
    }
};