zoukankan      html  css  js  c++  java
  • 数据结构——二叉树的遍历

    Description

    Download as PDF

    Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.

    This is an example of one of her creations:

                                        D
                                       / 
                                      /   
                                     B     E
                                    /      
                                   /         
                                  A     C     G
                                             /
                                            /
                                           F
    

    To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree).

    For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.

    She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).


    Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.

    However, doing the reconstruction by hand, soon turned out to be tedious.

    So now she asks you to write a program that does the job for her!

    Input Specification 

    The input file will contain one or more test cases. Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)

    Input is terminated by end of file.

    Output Specification 

    For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).

    Sample Input 

    DBACEGF ABCDEFG
    BCAD CBAD
    

    Sample Output 

    ACBFGED
    CDAB


    解题思路:用递归调用的形式,来根据前序,中序求后序,前序遍历的第一个数即是树的根,再将找出根在中序遍历的位置,放在P中,依次调用先序和后序,然后再寻找中序
    程序代码:
    #include <iostream>
    using namespace std;
    
    void build(int n,char * s1,char * s2,char * s)
    {
        if(n<=0)    return;
        int p=strchr(s2,s1[0])-s2;
        build(p,s1+1,s2,s);
        build(n-p-1,s1+p+1,s2+p+1,s+p);
        s[n-1]=s1[0];
    }
    int main()
    {
        char s1[30],s2[30],s[30];
        while(cin>>s1>>s2)
        {
            int n=strlen(s1);
            build(n,s1,s2,s);
            s[n]='';
            cout<<s<<"
    ";
        }
        return 0;
    }

    版权声明:此代码归属博主, 请务侵权!
  • 相关阅读:
    数组的应用:一。冒泡排序二。折半查找!二维数组的学习。
    break与continue,while 循环和一维数组的学习及作业
    for循环的应用:迭代法和穷举法
    循环
    称体重
    js js弹出框、对话框、提示框、弹窗总结
    windows 服务器开设端口
    SQL Server 数据库分离与附加(图文教程)
    ASP.NET MVC5 PagedList分页示例
    mvc 连接数据库但单复值得问题
  • 原文地址:https://www.cnblogs.com/www-cnxcy-com/p/4666389.html
Copyright © 2011-2022 走看看