好久没有写随笔了。。呵呵。。
呵呵 步ASP后尘 写他的题去。。。-_-!!!
看到一个题目 说是已知（input)一棵树的前序和中序遍历 要求输出后序遍历
我的算法很简单啦 就拿个字符串按照遍历的结构剪来剪去 呵呵 后来又想如果我要得到这棵树在内存中的状态呢？（也就是从上到下的长相） 于是添加了个东东 呵呵 随笔上来 各位见笑。。 呵呵
solution:
//by Optimistic
#include <iostream>
#include <string>
#include <math.h>
using namespace std;
int maxk;
string sa, sb;
char dst[1000];
int index[30];
void init()
{
//initiation
maxk = 0;
memset(dst, '^', sizeof(dst));
memset(index, 0, sizeof(index));
cout << "The PostOrder Of the tree:\n";
}
void cal_tree(string sa, string sb)
{
if(sb.length() == 0) return;
if(sb.length() == 1) {cout << sb;return;}
char x = sa[0];
int mid = sb.find(x);
string c = sb.substr(0, mid);
string d = sb.substr(mid+1);
cal_tree(sa.substr(1, c.length()), c);
cal_tree(sa.substr(1+c.length()), d);
cout << x;
}
void cal_BFStree(string sa, string sb, char * dst, int k, int pos)
{
if(k>maxk) maxk = k;
if(sb.length() == 0) return;
if(sb.length() == 1)
{
dst[(int)pow(2, k-1)-1+pos-1] = sb[0];
return;
}
char x = sa[0];
dst[(int)pow(2, k-1)-1+pos-1] = x;
int mid = sb.find(x);
string c = sb.substr(0, mid);
string d = sb.substr(mid+1);
cal_BFStree(sa.substr(1, c.length()), c, dst, k+1, 2*pos-1);
cal_BFStree(sa.substr(1+c.length()), d, dst, k+1, 2*pos);
}
void work()
{
cal_tree(sa, sb);
cal_BFStree(sa, sb, dst, 1, 1);
}
void output()
{
cout << endl;
int i, k=0;
cout << "The Tree in the RAM is like this:-) \n";
for(i=0; i<pow(2, sa.length()); i++)
{
cout << dst[i];
if(i==pow(2, k)-1) k++;
if(k>maxk) break;
}
cout << endl;
}
int main()
{
while(cin >> sa >> sb)
{
init();
work();
output();
}
return 0;
}
Sample Input
DBACEGF ABCDEFG
BCAD CBAD
Sample Output
DBACEGF ABCDEFG
The PostOrder Of the tree:
ACBFGED
The Tree in the RAM is like this:-)
DBEAC^G^^^^^^F^^
BCAD CBAD
The PostOrder Of the tree:
CDAB
The Tree in the RAM is like this:-)
BCA^^^D^
Original Problem Tree Recovery
Time Limit:1000MS Memory Limit:65536K
Total Submit:451 Accepted:325
Description
Little 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
The input 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
For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).
Sample Input
DBACEGF ABCDEFG
BCAD CBAD
Sample Output
ACBFGED
CDAB
Source
Ulm Local 1997