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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].
其实题目的意思就是相当于二叉树每一行的最右边的一个元素，那么简单，先bfs后取每一个数组的最后一位数就可以了，代码如下：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> rightSideView(TreeNode* root) { 13 int dep = -1; 14 bfs(root, dep + 1); 15 vector<int> res; 16 for(int i = 0; i < ret.size(); ++i){ 17 res.push_back(ret[i][ret[i].size() - 1]); 18 } 19 return res; 20 } 21 22 void bfs(TreeNode * root, int depth) 23 { 24 if(root == NULL) return; 25 if(depth < ret.size()){ 26 ret[depth].push_back(root->val); 27 }else{ 28 vector<int>tmp; 29 ret.push_back(tmp); 30 ret[depth].push_back(root->val); 31 } 32 if(root->left) 33 bfs(root->left, depth + 1); 34 if(root->right) 35 bfs(root->right, depth + 1); 36 } 37 private: 38 vector<vector<int>> ret; 39 };
LeetCode OJ：Binary Tree Right Side View（右侧视角下的二叉树）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4906486.html
