二叉树——建数字树，知先序、中序，输出后序

最新推荐文章于 2023-03-15 20:23:55 发布

QiujieDong

最新推荐文章于 2023-03-15 20:23:55 发布

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分类专栏： C&&C++ 文章标签：二叉树遍历

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本文介绍了一种从先序和中序遍历序列构造二叉树并获取后序遍历序列的方法。通过递归地构建二叉树，文章提供了一个具体的C++实现示例，帮助读者理解不同遍历方式之间的内在联系。

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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>
#include"cstdio"
#include"cstring"
#include"cstdlib"
using namespace std;
int  pre[1050],ins[1050];
int t;
struct node
{
    int data;
    node *lch,*rch;
};
node *build(int*pre ,int *ins,int n)
{
    if(n<=0) return NULL;
    node *p;
    p=(node*)malloc(sizeof(node));
    p->data=pre[0];
    int k;
    for(k=0;k<n;k++)
    {
        if(ins[k]==pre[0])
            break;
    }
    p->lch=build(pre+1,ins,k);
    p->rch=build(pre+k+1,ins+k+1,n-k-1);
    return p;
}
int num;
void houxu(node *p)
{
    if(p!=NULL)
    {
        houxu(p->lch);
        houxu(p->rch);
        if(num==0)
            {printf("%d",p->data);num++;}
        else
            printf(" %d",p->data);
    }
}
int main()
{
    while(~scanf("%d",&t))
    {
        num=0;
        memset(pre,0,sizeof(pre));
         memset(ins,0,sizeof(ins));
       for(int i=0;i<t;i++)
        scanf("%d",&pre[i]);
       for(int i=0;i<t;i++)
        scanf("%d",&ins[i]);
        node *tree;
        tree=(node*)malloc(sizeof(node));
        tree=build(pre,ins,t);
        houxu(tree);
        putchar('\n');
    }
    return 0;
}