hdu 1622 Trees on the level(二叉树的层次遍历)
题目链接:https://vjudge.net/contest/209862#problem/B
题目大意:
Trees on the level
Total Submission(s): 591 Accepted Submission(s): 200
This problem involves building and traversing binary trees.
Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes.
In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.
For example, a level order traversal of the tree
is: 5, 4, 8, 11, 13, 4, 7, 2, 1.
In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L's and R's where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.
All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.
思路:先建树,判断其能否构成二叉树,若能,则用bfs层次遍历该二叉树
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<queue>
using namespace std; const int maxn = + ; struct Node{
bool have_value;
int v;
Node* left, *right;
Node():have_value(false),left(NULL),right(NULL){}
}; Node* root; Node* newnode() { return new Node(); } bool failed;
void addnode(int v, char* s) {
int n = strlen(s);
Node* u = root;
for(int i = ; i < n; i++)
if(s[i] == 'L') {
if(u->left == NULL) u->left = newnode();
u = u->left;
} else if(s[i] == 'R') {
if(u->right == NULL) u->right = newnode();
u = u->right;
}
if(u->have_value) failed = true;
u->v = v;
u->have_value = true;
} void remove_tree(Node* u) {
if(u == NULL) return;
remove_tree(u->left);
remove_tree(u->right);
delete u;
} char s[maxn];
bool read_input() {
failed = false;
remove_tree(root);
root = newnode();
for(;;) {
if(scanf("%s", s) != ) return false;
if(!strcmp(s, "()")) break;
int v;
sscanf(&s[], "%d", &v);
addnode(v, strchr(s, ',')+);
}
return true;
} bool bfs(vector<int>& ans) {
queue<Node*> q;
ans.clear();
q.push(root);
while(!q.empty()) {
Node* u = q.front(); q.pop();
if(!u->have_value) return false;
ans.push_back(u->v);
if(u->left != NULL) q.push(u->left);
if(u->right != NULL) q.push(u->right);
}
return true;
} int main() {
vector<int> ans;
while(read_input()) {
if(!bfs(ans)) failed = ;
if(failed) printf("not complete\n");
else {
for(int i = ; i < ans.size(); i++) {
if(i != ) printf(" ");
printf("%d", ans[i]);
}
printf("\n");
}
}
return ;
}
2018-04-10
hdu 1622 Trees on the level(二叉树的层次遍历)的更多相关文章
- 【二叉树】hdu 1622 Trees on the level
[题意] 给定一棵树每个结点的权重和路径(路径用LR串表示),输出这棵树的层次遍历 [思路] 注意输入输出,sscanf用来格式化地截取需要的数据,strchr来在字符串中查找字符的位置 [Accep ...
- hdu 1622 Trees on the level
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1622 小白书上的题... #include<algorithm> #include< ...
- UVA.122 Trees on the level(二叉树 BFS)
UVA.122 Trees on the level(二叉树 BFS) 题意分析 给出节点的关系,按照层序遍历一次输出节点的值,若树不完整,则输出not complete 代码总览 #include ...
- Trees on the level UVA - 122 (二叉树的层次遍历)
题目链接:https://vjudge.net/problem/UVA-122 题目大意:输入一颗二叉树,你的任务是按从上到下,从左到右的顺序输出各个结点的值.每个结点都按照从根节点到它的移动序列给出 ...
- LeetCode 102. Binary Tree Level Order Traversal 二叉树的层次遍历 C++
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, ...
- LeetCode 102. 二叉树的层次遍历(Binary Tree Level Order Traversal) 8
102. 二叉树的层次遍历 102. Binary Tree Level Order Traversal 题目描述 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 每 ...
- lintcode : 二叉树的层次遍历II
题目 二叉树的层次遍历 II 给出一棵二叉树,返回其节点值从底向上的层次序遍历(按从叶节点所在层到根节点所在的层遍历,然后逐层从左往右遍历) 样例 给出一棵二叉树 {3,9,20,#,#,15,7}, ...
- lintcode : 二叉树的层次遍历
题目 二叉树的层次遍历 给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问) 样例 给一棵二叉树 {3,9,20,#,#,15,7} : 3 / \ 9 20 / \ 15 7 返回他的分层遍历 ...
- LintCode 二叉树的层次遍历 II
中等 二叉树的层次遍历 II 查看执行结果 42% 通过 给出一棵二叉树,返回其节点值从底向上的层次序遍历(按从叶节点所在层到根节点所在的层遍历,然后逐层从左往右遍历) 您在真实的面试中是否遇到过这个 ...
随机推荐
- POJ1236 Network of Schools【强连通】
题意: N(2<N<100)各学校之间有单向的网络,每个学校得到一套软件后,可以通过单向网络向周边的学校传输,问题1:初始至少需要向多少个学校发放软件,使得网络内所有的学校最终都能得到软件 ...
- mongodb系列~mongodb数据迁移
一 简介:今天来聊聊mongo的数据迁移二 迁移 1 具体迁移命令 nohup mongodump --port --db dbname --collection tablename --qu ...
- linux笔记_day06
1.用户:表示符,凭证 2.用户组:表示符 进程也是有属主和属组的 安全上下文(secure context): 用户:UID,/etc/pawwd 组:GID ,/etc/group 影子口令: 用 ...
- python 历险记(二)— python 的面向对象
目录 前言 类和对象 如何定义和实例化类? 如何定义和使用属性? 什么是方法? 静态方法和普通方法 构造函数该怎么写? str 方法怎么写? 多态是什么? 继承性和 java 是一样的吗? 父类和子类 ...
- Linux内核驱动之延时 【转】
转自:http://blog.chinaunix.net/uid-24219701-id-3288103.html jiffies 计数器 定时器中断由系统定时硬件以规律地间隔产生; 这个间隔在启动 ...
- sqlserver中将查询结果拼接成字符串
#for xml path(param)--将查询结果以xml格式输出 select id,name from table1 for xml path --id和name为table1的真实字段 - ...
- Innodb ,MyISAM
1. InnoDB不支持FULLTEXT类型的索引. 2. InnoDB 中不保存表的具体行数,也就是说,执行select count(*) from table时,InnoDB要扫描一遍整个表来计算 ...
- js 、c# 编码解码
escape不编码字符有69个:*,+,-,.,/,@,_,0-9,a-z,A-Z encodeURI不编码字符有82个:!,#,$,&,',(,),*,+,,,-,.,/,:,;,=,?,@ ...
- spring boot 中的热部署
<plugin> <groupId>org.springframework.boot</groupId> <artifactId>sprin ...
- spring事物回滚遇到的问题
在service层使用声明式事务添加@Transactional(rollbackFor = Exception.class)注解 多个方法进行数据库操作,执行失败则隐式的回滚事务,但是已经成功的发方 ...