码迷,mamicode.com
首页 > 其他好文 > 详细

【Leetcode】【Medium】Binary Tree Right Side View

时间:2015-05-22 09:22:05      阅读:84      评论:0      收藏:0      [点我收藏+]

标签:

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].

 

解题思路:

这题和上一题Binary Tree Zigzag Level Order Traversal很相似,都需要按层遍历二叉树;

不同的是,由于Binary Tree Zigzag Level Order Traversal需要Z型遍历,需要用到“先入后出”的方法,因此用栈stack来实现;

而本题,需要每层由右向左遍历,因此用到“先进先出”,所以使用queue来实现较为方便;

 

需要两个queue,一个保存当前层的结点,另一个保存下一层的结点;

每层结点queue中第一个值,就是最右端值,输出这个值。

代码:

 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         vector<int> ret;
14         queue<TreeNode*> cur_layer;
15         cur_layer.push(root);
16         queue<TreeNode*> next_layer;
17         if (!root)
18             return ret;
19         
20         while (!cur_layer.empty()) {
21             ret.push_back(cur_layer.front()->val);
22             while (!cur_layer.empty()) {
23                 TreeNode* node = cur_layer.front();
24                 cur_layer.pop();
25                 if (node->right)
26                     next_layer.push(node->right);
27                 if (node->left)
28                     next_layer.push(node->left);
29             }
30             swap(cur_layer, next_layer);
31         }
32         
33         return ret;
34     }
35 };

 

【Leetcode】【Medium】Binary Tree Right Side View

标签:

原文地址:http://www.cnblogs.com/huxiao-tee/p/4521369.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!