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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; } };
LeetCode OJ：Binary Tree Right Side View
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原文地址：http://www.cnblogs.com/csuer/p/4441170.html
