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。
1 class Solution { 2 public: 3 void help(vector<int>& res, TreeNode* root, int depth) 4 { 5 if (!root) return; 6 if (res.size() < depth + 1) 7 res.push_back(root->val); 8 help(res, root->right, depth + 1); 9 help(res, root->left, depth + 1); 10 } 11 vector<int> rightSideView(TreeNode* root) { 12 vector<int> res; 13 help(res, root, 0); 14 return res; 15 } 16 };