zoukankan      html  css  js  c++  java
  • POJ2255Tree Recovery

    2255:Tree Recovery

    题目分析:使用前序序列和中序序列构造后序序列的题。当时面微软实习生的时候还问过这个。

    前序的第一个节点就是后序的最后一个节点,且该节点把中序序列划分成左右两个子树。中序的这两个子树对应着前序的两个子树,他们的长度分别相等。

    代码如下:

    #include <iostream>
    #include <string>
    using namespace std;
    string getPost(string pre,string in){
        if(pre.length() == 0)return "";//注意边界条件
        if(pre.length() == 1)return pre;
        char root = pre[0];
        int len = in.find(root);
        return getPost(pre.substr(1,len),in.substr(0,len)) + getPost(pre.substr(1+len),in.substr(1+len)) + root;
    }
    int main(){
        string pre,in;
        while(cin>>pre){
            cin>>in;
            cout<<getPost(pre,in)<<endl;
        }
        return 0;
    }
    题目:

    时间限制: 1000ms 内存限制: 65536kB
    描述
    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!


    输入
    The input 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.


    输出
    For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).
    样例输入
    DBACEGF ABCDEFG
    BCAD CBAD
    样例输出
    ACBFGED
    CDAB

  • 相关阅读:
    【转】验证实例名称是否正确并且 SQL Server 已配置为允许远程连接。
    【转】在控制台、WinForm项目中的嵌入mdf文件的烦恼
    【转】ADB server didn't ACK
    【转】android中AVD的使用
    【转】你不可以不知道的findViewById
    【转】Android中的view
    【转】c# 如何获取项目的根目录
    TYVJ 1452 最大权闭合图
    NOIP 最后的总结
    POJ 2396 有上下界的可行流
  • 原文地址:https://www.cnblogs.com/zhuyuanhao/p/3262878.html
Copyright © 2011-2022 走看看