求后序

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.

Input
The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.
Output
For each test case print a single line specifying the corresponding postorder sequence.
Sample Input
9
1 2 4 7 3 5 8 9 6
4 7 2 1 8 5 9 3 6
Sample Output
7 4 2 8 9 5 6 3 1

题意：给出一棵二叉树的前序和中序，输出后序

思路：和求前序的思路差不多，利用每棵树的前序第一个为根节点，找到中序中与之对应的值，把中序分为左树和右树，把根节点保存（保存是在两个递归调用之后的）。

代码：

#include"stdio.h"
#include"string.h"
char a[1010],b[1010],s[1010];
int book[1010],k;
void dfs(int l1,int r1,int l2,int r2)
{
    int root,left,right;
    root=book[int(a[l1]-'A'+1)];
    left=root-l2;
    right=r2-root;
    if(left)
    dfs(l1+1,l1+root-1,l2,root-1);//左树
    if(right)
    dfs(l1+left+1,r1,root+1,r2);//右树
    s[k++]=a[l1];
}
int main()
{
    while(~scanf("%s %s",a,b))
    {
        int i,l;
        k=0;
        l=strlen(a);
        for(i=0;i<l;i++)
        book[int(b[i]-'A'+1)]=i;
        dfs(0,l-1,0,l-1);
        for(i=0;i<l;i++)
		printf("%c",s[i]);
        printf("\n");
    }
    return 0;
}