POJ 2255. Tree Recovery
Tree Recovery
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11939 | Accepted: 7493 |
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:
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!
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.
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
网上大量的人在秀数据结构基础，栈构造树都出来了呵呵。
其实这题很简单也很水，根据前序遍历、中序遍历输出后序遍历。那么根据三种遍历的定义，在前序遍历中的第一个就是根节点，在中序遍历里找到它，左边的就是左子树，右边的就是右子树，以此类推。因此，只要递归地分割字符串就好了。可以在POJ上看到 05 - 07 年提交的程序内存占用都很低，反正我是做不到，可能那时的数据也很水吧。以下程序 124ms。
1 #include <stdio.h> 2 char in[27],pre[27],*pr; 3 void maketree(char *in) 4 { 5 char *mid,tmp; 6 if(*in=='\0') 7 return; 8 //mid = strchr(in,*pr++); 9 for(mid=in;*mid!=*pr;) 10 mid++; 11 pr++; 12 13 tmp = *mid; 14 *mid = '\0'; 15 mid++; 16 maketree(in); 17 maketree(mid); 18 putchar(tmp); 19 } 20 int main() 21 { 22 while(scanf("%s%s", pre, in) == 2) { 23 pr=pre; 24 maketree(in); 25 putchar('\n'); 26 } 27 return 0; 28 }
我的递归里面因为不是用下标处理的，也没在参数里加长度来判断，所以显得麻烦点。如果加了长度判断，则几行就搞定了（代码来自SCNU变态的小liming）：
1 void solve(char*a,char*b,int L){ 2 char*i; 3 if(L){ 4 i=strchr(b,*a); 5 solve(a+1,b,i-b); 6 solve(a-b+i+1,i+1,b+L-i-1); 7 putchar(*a); 8 } 9 }
