首页 > ACM题库 > HDU-杭电 > HDU 1710 Binary Tree Traversals-递归和分治-[解题报告] C++
2013
12-21

HDU 1710 Binary Tree Traversals-递归和分治-[解题报告] C++

Binary Tree Traversals

问题描述 :

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.

输入:

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.

输出:

For each test case print a single line specifying the corresponding postorder sequence.

样例输入:

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

样例输出:

7 4 2 8 9 5 6 3 1

题意:根据前序和中序写出后序  前序:1 2 4 7 3 5 8 9 6 中序:4 7 2 1 8 5 9 3 6 求出后序:7 4 2 8 9 5 6 3 1

首先得知道是如何前序遍历、中序遍历、后序遍历的,自己上网查下,我在这里就不多说了

思路:第一步:根据前序可知根节点为1;第二步:根据中序可知4 7 2为根节点1的左子树和8 5 9 3 6为根节点1的右子树;第三步:递归实现,把4 7 2当做新的一棵树和8 5 9 3 6也当做新的一棵树;第四步:在递归的过程中输出后序。

代码实现:

//根据前序和中序遍历写出后序遍历
#include<iostream>
using namespace std;
int t1[1001],t2[1001];
void sousuo(int a,int b,int n,int flag)
{
    
    if(n==1)//如果存在左子树或右子树就直接输出
    {
        printf("%d ",t1[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;//继续罚分为左子树和右子树
    for(i=0;t1[a]!=t2[b+i];i++) ;//找到罚分点也就是根节点
    sousuo(a+1,b,i,0);//左子树的遍历
    sousuo(a+i+1,b+i+1,n-i-1,0);//右子树的遍历
    if(flag==1)//最原始的跟节点
        printf("%d",t1[a]);
    else//一般的根节点
        printf("%d ",t1[a]);
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&t1[i]);//t1中存的是前序
        for(i=1;i<=n;i++)//t2中存的中序
            scanf("%d",&t2[i]);
        sousuo(1,1,n,1);
        printf("\n");
    }
    return 0;
}

这是根据前序和中序写出后序,我现在把题意变成根据后序和中序写出前序 后序:7 4 2 8 9 5 6 3 1 中序:4 7 2 1 8 5 9 3 6 求出前序:1 2 4 7 3 5 8 9 6
代码实现:

//根据后序和中序遍历写出前序
#include<iostream>
using namespace std;
int t1[1001],t2[1001];
void sousuo(int a,int b,int n,int flag)
{
    int i;
    if(n==1)//存在左子树或右子树,进行遍历即可
    {
       printf(" %d",t1[a]);
       return ;
    }
    else if(n<=0)//不存在左子树或者右子树,则返回上一层
        return ;
    if(flag==1)//最原始的根节点
        printf("%d",t1[a]);
    else //一般的根节点
        printf(" %d",t1[a]);
    for(i=0;t1[a]!=t2[b+i];i++) ;//找出根节点
    sousuo(a-n+i,b,i,0);//左子树的遍历
    sousuo(a-1,b+i+1,n-i-1,0);//右子树的遍历
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&t1[i]);//t1中存的是后序
        for(i=1;i<=n;i++)
            scanf("%d",&t2[i]);//t2中存的是中序
        sousuo(n,1,n,1);//因为后序最后遍历的是根节点,所以这里和前面的开始点不同,注意一下
        printf("\n");
    }
    return 0;
}

 

 

解题报告转自:http://www.cnblogs.com/jiangjing/archive/2013/01/14/2860163.html


  1. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }

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