好久沒有寫隨筆了。。呵呵。。
呵呵 步ASP后塵 寫他的題去。。。-_-!!!
看到一個(gè)題目 說是已知（input)一棵樹的前序和中序遍歷 要求輸出后序遍歷
我的算法很簡單啦 就拿個(gè)字符串按照遍歷的結(jié)構(gòu)剪來剪去 呵呵 后來又想如果我要得到這棵樹在內(nèi)存中的狀態(tài)呢？（也就是從上到下的長相） 于是添加了個(gè)東東 呵呵 隨筆上來 各位見笑。。 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