Binary Tree Traversals（HDU1710）二叉树的简单应用

 

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2681    Accepted Submission(s): 1178

Problem Description

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

 

 

#include<iostream>

using namespace std;



typedef struct tree// typedef定义类型,这里为结构体型;

{

    int num;

    struct tree *left;

    struct tree *right;

};



tree *root;



tree *creat(int *a,int *b,int n)//用前序历遍和中序历遍得到的数据，确定唯一树



{

    tree *k;

    int i;

    for(i=0;i<n;i++)

    {

        if(*a==b[i])//当在中序历遍中找到了根节点后

        {

            k=(tree *)malloc(sizeof(tree));

            k->num=b[i];

            k->left=creat(a+1,b,i);//中序历遍中在根节点左边的都是左子树上的

            k->right=creat(a+i+1,b+i+1,n-i-1);//在根节点右边的，都是右子树上的，右子树需要从i+1开始；

            //因为他的根的左半只有i个数，加上自己所有就要把指针指向a+i+1的地方了，



//            printf("%d\n",h->data );直接输出于是后续，不过要判断不成立的情况，所以不行不过我想可以用数组装起来;

            return k;

        }

    }

    return NULL;//没有对应找到的话，就返回NULL

}



void houxu(tree *h)

{

    if(h!=NULL)

    {

        houxu(h->left);

        houxu(h->right);

        if(root==h)//后序历遍最后历遍根节点

        {

            printf("%d\n",h->num);

        }

        else

        {

            printf("%d ",h->num);

        }

    }

}

int main()

{

    tree *h;

    int a[1001],b[1001],n,i;

    while(scanf("%d",&n)!=EOF&&n)

    {

        for(i=0;i<n;i++)

            scanf("%d",&a[i]);

        for(i=0;i<n;i++)

            scanf("%d",&b[i]);

        h=creat(a,b,n);

        root=h;

        houxu(h);

    }

}