Topic:
Description
Little Valentine liked playing with binary trees very much. Her favorite game is constructing randomly looking binary trees with capital letters in the nodes.
This is a example of one of her creations:
D / / B E /\ / \ \ A C G / / F
To record hers trees for the future generations, she wrote down the strings for each tree:a preorder traversal (root, left Sub Tree, right subtree) and a inorder traversal (left subtree, root, right subtree).
For the tree drawn above the preorder traversal are 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 I T).
Now, years later, looking again in the strings, she realized that reconstructing the trees is indeed possible, but only B Ecause she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned off to be tedious.
So now she asks you to write a program this does the job for her!
Input SpecificationThe input file is contain one or more test cases. Each test case consists of one line containing the strings Preord and Inord, representing the preorder traversal and Inord ER traversal of a binary tree. Both strings consist of unique capital letters. (Thus they is not longer than characters.)
Input is terminated by end of file.
Output SpecificationFor each test case, recover Valentine's binary tree and print one line containing the tree's Postorder traversal (left Sub Tree, right subtree, root). The main idea: to give you the first sequence of the tree and the sequence traversal, you have to output post-order traversal. Problem-solving ideas: Give a structure with a tree root Saozi right child number. Use recursion to build up the tree. In the output. Code:
1#include <iostream>2#include <cstring>3 using namespacestd;4 Const intmaxn= -;5 structShu6 {7 CharA;8shu*l;9shu*R;Ten }B[MAXN]; Oneshu* Goujianshu (shu* D,Char* Q,Char* W,intN) A { - inti; - if(n==0) the returnNULL; - for(i=0; i<n;i++) - { - if(q[0]==W[i]) + { -d->a=q[0]; +D->l=goujianshu (d+1, q+1, w,i); AD->r=goujianshu (d+1+i,q+i+1, w+i+1, n-i-1); at } - } - returnD; - } - voidHouxu (shu*d) - { in if(d==NULL) - return; toHouxu (d->l); +Houxu (d->R); -cout<<Char(d->a); the } * intMain () $ {Panax Notoginseng CharE[MAXN],T[MAXN]; - while(cin>>e>>t) the { + Goujianshu (B,e,t,strlen (e)); A Houxu (b); thecout<<Endl; + } - return 0; $}
Huas Summer training#2 D