模板--二叉树

爱被打了一巴掌 2022-06-11 04:21 124阅读 0赞

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

思路：  
  
大致题意，已知前序和中序，求后序。根据二叉树的构建思想即可。  
  
代码：

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    using namespace std;   int t;
    struct treenote
    {
        int elem;
        treenote *lchild;
        treenote *rchild;
    };
    treenote *gettree(int  *preorder,int  *inorder ,int len)
    {
        if(len<=0)
            return NULL;
        treenote *note =new treenote;
        note ->elem =*preorder;
    int     root=0;
       while(root<len)
       {
           if(*preorder==inorder[root])
            break;
           root++;
       }
        note->lchild=gettree(preorder+1,inorder,root);
        note->rchild=gettree(preorder+1+root,inorder+root+1,len-root-1);
       len==t? cout<<note->elem :cout<<note->elem<<" ";
        return note;
    
    }
    int main()
    {
    
        int preorder[1020],inorder[1020];
        while (cin>>t&&(1<=t<=1000))
    
       {
           for(int i=0;i<t;i++)
            cin>>preorder[i];
           for(int i=0;i<t;i++)
            cin>>inorder[i];
        gettree(preorder, inorder, t);
        cout<<endl;
       }
        return 0;
    
    }