PAT1020. Tree traversals

Suppose the keys in a binary tree is distinct positive integers. Given the Postorder and inorder traversal sequences, you is supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains the one test case.  For each case, the first line gives a positive integer N (<=30) and the total number of nodes in the binary tree.  The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line is separated by a space.

Output Specification:

For each test case, the print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must is separated by exactly one space, and there must is no extra space at the end of the line.

Sample Input:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2
Thought: The sequence of the sequence  and then the hierarchy of sequential traversal  is very simple problem. It took a long time because of punctuation, and if it was followed by a punctuation mark.

1#include <cstdio>2#include <cstring>3#include <queue>4#include <algorithm>5 using namespacestd;6 Const intmaxn= -;7 structnode8 {9     intdata;TenNode *Lchild; OneNode *Rchild; A }; - int inch[MAXN],POST[MAXN]; - intN; theNode * Create (intPostl,intPOSTR,intInL,intInR) - { -     if(postl>postr) -       returnNULL; +Node *root=Newnode; -Root->data=Post[postr]; +      intK; A       for(k=inl;k<=inr;k++) at      { -          if(inch[k]==Post[postr]) -          { -               Break; -          } -      } in      intnumleft=k-InL; -Root->lchild=create (postl,postl+numleft-1, inl,k-1); toRoot->rchild=create (postl+numleft,postr-1, K +1, InR); +      returnRoot; - }  the  * //perform a hierarchical traversal $ intnum=0;Panax Notoginseng voidBFS (Node *root) - { thequeue<node* >Q; + Q.push (root); A      while(!q.empty ()) the     { +Node *now=Q.front (); - Q.pop ();  $printf"%d",now->data); $num++; -         if(num<N) -printf" ");  the         if(now->lchild!=NULL) -Q.push (now->lchild);Wuyi         if(Now->rchild!=null)//because it takes a long time to add a punctuation mark here.  theQ.push (now->rchild); -     } Wu } -  About intMainintargcChar*argv[]) $ { -  -scanf"%d",&n); -      for(intI=0; i<n;i++) Ascanf"%d",&post[i]); +      for(intI=0; i<n;i++) thescanf"%d",&inch[i]); -node* Root=create (0, N-1,0, N-1); $ BFS (root); the     return 0; the}

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PAT1020. Tree traversals