數據結構實驗之二叉樹四：還原二叉樹

TimeLimit: 1000MS Memory Limit: 65536KB

SubmitStatistic

PRoblem Description

給定一棵二叉樹的先序遍歷序列和中序遍歷序列，要求計算該二叉樹的高度。

Input

輸入數據有多組，每組數據第一行輸入1個正整數N(1 <= N <= 50)為樹中結點總數，隨后2行先后給出先序和中序遍歷序列，均是長度為N的不包含重復英文字母(區分大小寫)的字符串。

Output

 輸出一個整數，即該二叉樹的高度。

Example Input

9

ABDFGHIEC

FDHGIBEAC

Example Output

5

Hint

Author

 xam

#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 last(tree*root){   if(root)   {      last(root->l);      last(root->r);      cout<<root->data;   }}int high(tree*root){   if(!root)   return 0;   else   return max(high(root->l),high(root->r))+1;}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;   while(cin>>o)   {    int len;   scanf("%s%s",xian,zhong);   len = strlen(zhong);   tree* root;   root = huifu2(xian,zhong,len);   int u = high(root);   cout<<u<<endl;   }}/***************************************************User name: jk160505徐紅博Result: AcceptedTake time: 0msTake Memory: 168KBSubmit time: 2017-02-07 15:58:38****************************************************/