A

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

91 2 4 7 3 5 8 9 64 7 2 1 8 5 9 3 6

Sample Output

7 4 2 8 9 5 6 3 1

#include<stdio.h>int A[1000],B[1000];void houxu(int a,int b,int n,int q){    if(n==1)//如果存在左子树或右子树就直接输出    {        printf("%d ",A[a]);        return ;    }    else if(n<=0)//如果不存在左子树或右子树就返回上一层        return ;    int i;    for(i=0;A[a]!=B[b+i];i++);//找到根节点，即为前序序列的第一个        houxu(a+1,b,i,0);//左子树的遍历        houxu(a+i+1,b+i+1,n-i-1,0);//右子树的遍历    if(q==1)//最原始的根节点            printf("%d",A[a]);    else        printf("%d ",A[a]);}int main ( ){    int i,n;    while(scanf("%d",&n)!=EOF)    {    for(i=1;i<=n;i++)        scanf("%d",&A[i]);//存放前序序列    for(i=1;i<=n;i++)        scanf("%d",&B[i]);//存放中序序列    houxu(1,1,n,1);    printf("\n");    }    return 0;}

上面题目是已知前序序列和中序序列求后续序列。

这里顺便拓展下已知后续序列和中序序列求前序序列

//知道后序和中序求前序#include<stdio.h>int A[1000],B[1000];void qianxu(int a,int b,int n,int q){    if(n==1)//如果存在左子树或右子树就直接输出    {        printf(" %d",A[a]);        return ;    }    else if(n<=0)//如果不存在左子树或右子树就返回上一层        return ;    if(q==1)//最原始的根节点            printf("%d",A[a]);    else        printf(" %d",A[a]);            int i;    for(i=0;A[a]!=B[b+i];i++);//找到根节点，即为后序序列的最后一个        qianxu(a-n+i,b,i,0);//左子树的遍历        qianxu(a-1,b+i+1,n-i-1,0);//右子树的遍历}int main ( ){    int i,n;    while(scanf("%d",&n)!=EOF)    {    for(i=1;i<=n;i++)        scanf("%d",&A[i]);//存放后序序列    for(i=1;i<=n;i++)        scanf("%d",&B[i]);//存放中序序列    qianxu(n,1,n,1);//因为后序最后遍历的是根节点    printf("\n");    }    return 0;}

递归思想很重要。