初次做二叉树的题目--给定中序和前序求后序

Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear `tree in-order' and `tree pre-order' notations.

Your job is to create the `tree post-order' notation of a cow's heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.

Here is a graphical representation of the tree used in the sample input and output:

C / \ / \ B G / \ / A D H / \ E F

The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree.

The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree.

The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root.

PROGRAM NAME: heritageINPUT FORMAT

 Line 1: The in-order representation of a tree. 

Line 2: The pre-order representation of that same tree. 

SAMPLE INPUT (file heritage.in)

ABEDFCHGCBADEFGHOUTPUT FORMAT

A single line with the post-order representation of the tree.

SAMPLE OUTPUT (file heritage.out)

AEFDBHGC 

思路：这题主要用到递归的思想代码：

#include<stdio.h>#include<string.h>#include<malloc.h>char root;FILE *fin,*fout;char pre_order[28],in_order[28];struct BNode{     char data;     BNode * ld,* rd;};int pos=0;void createtree(BNode * &b,int first,int last) //这里用了引用类型{     int i;    if(first>last)         return ;     b=(BNode *)malloc(sizeof(BNode));     for(i=first;i<=last;i++)         if(in_order[i]==pre_order[pos])             break;     b->data=pre_order[pos];    pos++; 

    b->ld=NULL;

    b->rd=NULL;     createtree(b->ld,first,i-1);    createtree(b->rd,i+1,last);}void post_traversal(BNode *b){    if(b->ld)     post_traversal(b->ld);     if(b->rd)         post_traversal(b->rd);     fprintf(fout,"%c",b->data);}int main(){    int len;     BNode *BTree;     fin=fopen("heritage.in","r");     fout=fopen("heritage.out","w");     fscanf(fin,"%s%s",in_order,pre_order);     len=strlen(in_order);     createtree(BTree,0,len-1);     post_traversal(BTree);     fprintf(fout,"\n");     return 0;}

Here are the test data inputs:

------- test 1 -------
ABEDFCHG
CBADEFGH
------- test 2 -------
F
F
------- test 3 -------
BCAD
ABCD
------- test 4 -------
GOLEAFS
SFAELOG
------- test 5 -------
GSHBAQTPM
ABGHSPQTM
------- test 6 -------
AUBYCVDZEWFXGTH
ZYUABVCDXWEFTGH
------- test 7 -------
ABDCJHKILMNPOQFEGRS
ABCDEFHJIKLMNOPQGRS
------- test 8 -------
GFDIHKLJMBNESRTPOQAUCWVZYX
ABDFGHIJKLMENOPRSTQCUVWXYZ
------- test 9 -------
EHGDIFJLKMBNCOQSPRAWUXZYTV
ABDEGHFIJKLMCNOPQSRTUWXYZV

Keep up the good work!