求二叉樹的層次遍歷

TimeLimit: 1000MS Memory Limit: 65536KB

SubmitStatistic

PRoblem Description

已知一顆二叉樹的前序遍歷和中序遍歷，求二叉樹的層次遍歷。

Input

輸入數(shù)據(jù)有多組，輸入T，代表有T組測試數(shù)據(jù)。每組數(shù)據(jù)有兩個長度小于50的字符串，第一個字符串為前序遍歷，第二個為中序遍歷。

Output

每組輸出這顆二叉樹的層次遍歷。

Example Input

2

abc

bac

abdec

dbeac

Example Output

abc

abcde

Hint

Author

fmh

#include<string.h>#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<queue>#include<iostream>using namespace std;typedef struct node{   char data;   struct node*l;   struct node*r;}tree;void huifu(char *zhong,char *hou,int len){  if(len==0)  return ;  tree *p = new tree;  p->data = *(hou+len-1);  int i = 0;  for(;i<len;i++)  if(zhong[i]==*(hou+len-1))  {     break;  }   cout<<p->data;  huifu(zhong,hou,i);  huifu(zhong+i+1,hou+i,len-i-1);  return ;}tree* huifu2(char *xian,char *zhong,int len){   tree*head =new tree;   if(len==0)   return NULL;   head->data = *(xian);   int i = 0;   for(;i<len;i++)   {       if(zhong[i]==*xian)       break;   }   head->l = huifu2(xian+1,zhong,i);   head->r = huifu2(xian+i+1,zhong+i+1,len-i-1);   return head;}void ccout(tree*root){    queue<tree*>q;    tree*p =NULL;    if(root)    {      q.push(root);    }    while(!q.empty())    {        p = q.front();        q.pop();        cout<<p->data;        if(p->l)        {           q.push(p->l);        }        if(p->r)        {           q.push(p->r);        }    }}int main(){   char hou[102],zhong[102],xian[102];   int o;   cin>>o;   while(o--)   {    int len;   scanf("%s%s",xian,zhong);   len = strlen(zhong);   tree* root;   root = huifu2(xian,zhong,len);   ccout(root);   cout<<endl;   }}/***************************************************User name: jk160505徐紅博Result: AcceptedTake time: 0msTake Memory: 164KBSubmit time: 2017-02-07 15:42:24****************************************************/