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  • POJ2255Tree Recovery

    Description

    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

    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. 

    Output

    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

    法一:非递归
    View Code
     1 #include <iostream>
     2 #include <cstring>
     3 #include <string>
     4 using namespace std;
     5 
     6 string s1,s2;
     7 
     8 typedef struct node{
     9     int pt;
    10     char x;
    11     struct node *left;
    12     struct node *right;
    13 }node;
    14 
    15 void postorder(node *root){
    16     if(root==NULL) return;
    17     postorder(root->left);
    18     postorder(root->right);
    19     cout<<root->x;
    20 }
    21 
    22 int main()
    23 {
    24     int i,j;
    25 
    26     while(cin>>s1>>s2){
    27         node *root=new(node);
    28             root->left=NULL;
    29             root->right=NULL;
    30             root->x=s1[0];
    31             root->pt=0;
    32             for(j=0;j<(int)s2.length();j++){
    33                 if(s1[0]==s2[j])
    34                     break;
    35             }
    36         root->pt=j;
    37 
    38         for(i=1;i<(int)s1.length();i++){
    39             for(j=0;j<(int)s2.length();j++){
    40                 if(s1[i]==s2[j])
    41                     break;
    42             }
    43             node *p=root, *pre=NULL;
    44 
    45             while(p){
    46                 pre=p;
    47 
    48                 if(j<p->pt){
    49                     p=p->left;
    50                 }else{
    51                     p=p->right;
    52                 }
    53 
    54             }
    55             node *t=new(node);
    56             t->left=NULL;
    57             t->right=NULL;
    58             t->x=s1[i];
    59             t->pt=j;
    60             if(j<pre->pt){
    61                 pre->left=t;
    62             }else{
    63                 pre->right=t;
    64             }
    65         }
    66         postorder(root);
    67         cout<<endl;
    68     }
    69     return 0;
    70 }

    法二:递归

    View Code
     1 #include<stdio.h>
     2 #include<string.h>
     3 void build(int n,char*s1,char*s2)
     4 {
     5     if(n<=0)return ;
     6     int p=strchr(s2,s1[0])-s2;
     7     build(p,s1+1,s2);
     8     build(n-p-1,s1+p+1,s2+p+1);
     9     printf("%c",s1[0]);
    10 }
    11 int main()
    12 {
    13     char s1[30],s2[30],ans[30];
    14     memset(s1,0,sizeof(s1));
    15     memset(s2,0,sizeof(s2));
    16     while(scanf("%s %s",s1,s2)!=EOF)
    17     {
    18         int n=strlen(s1);
    19         build(n,s1,s2);
    20         ans[n]='\0';
    21         printf("\n");
    22     }
    23     return 0;
    24 }
    ==================================================

    作者: Panderen

    博客: http://panderen.cnblogs.com

    签名: 机会总是为有准备的人而准备的!

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  • 原文地址:https://www.cnblogs.com/panderen/p/2443121.html
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