2015 HUAS Summer Training#2 D
2015-07-24 21:02
369 查看
题目:
DescriptionLittle Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:
D / \ / \ B E / \ \ / \ \ A C G / / F
To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree).
For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).
Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!
Input Specification
The input file will contain one or more test cases. Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)Input is terminated by end of file.
Output Specification
For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).题目大意:给你树的先序遍历和中序遍历,要你输出后序遍历。
解题思路:给个结构体有树的根 左子树和右子数。用递归把树构建起来。在输出。
代码:
#include<iostream> #include<cstring> using namespace std; const int maxn=30; struct shu { char a; shu* l; shu* r; }b[maxn]; shu* goujianshu(shu* d,char* q,char* w,int n) { int i; if(n==0) return NULL; for(i=0;i<n;i++) { if(q[0]==w[i]) { d->a=q[0]; d->l=goujianshu(d+1,q+1,w,i); d->r=goujianshu(d+1+i,q+i+1,w+i+1,n-i-1); } } return d; } void houxu(shu* d) { if(d==NULL) return; houxu(d->l); houxu(d->r); cout<<char(d->a); } int main() { char e[maxn],t[maxn]; while(cin>>e>>t) { goujianshu(b,e,t,strlen(e)); houxu(b); cout<<endl; } return 0; }
相关文章推荐
- 2015 HUAS Summer Training#2 C
- 狡猾的百度安全组件、百度安全防护服务
- 2015 HUAS Summer Training#2 B
- hdu 5307 He is Flying 2015 Multi-University Training Contest 2 快速傅里叶变换
- Hdu 5302 Connect the Graph 2015 Multi-University Training Contest 2
- 利用standford-nlp库实现Naive Bayes文本分类系统
- 【CODEFORCES】 A. Dreamoon and Stairs
- 更换STM32芯片类型引起keil下载Error Flash download failed-Cortex-M3问题的解决
- NSURLErrorDomain错误码
- HDU 5305 Friends 2015 Multi-University Training Contest 2 1006
- failed to open stream: HTTP wrapper does not support writeable connections
- 2015 HUAS Summer Training#2 A
- zoj1713 Haiku Review
- 使用CURL出现certificate verify failed错误的解决方法
- Assertion failed (ni > 0 && ni == ni1) in collectCalibrationData
- rails join left结合使用
- __attribute__((constructor)) 修饰的函数在main函数之前执行
- 朴素贝叶斯算法(naive Bayes algorithm)
- Android控件使用手册:使用WebView打开http://www.baidu.com
- 邮件部署