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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].
/** * 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: void dfs(TreeNode* root,vector<int>& res,int& res_size,int depth){ if(!root) return ; if(depth==res_size){ res_size++; res.push_back(root->val); } dfs(root->right,res,res_size,depth+1); dfs(root->left,res,res_size,depth+1); } vector<int> rightSideView(TreeNode* root) { vector<int> res; int res_size =0; dfs(root,res,res_size,0); return res; } };
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原文地址：http://www.cnblogs.com/zengzy/p/5054930.html
