数据结构实验之二叉树四:还原二叉树
TimeLimit: 1000MS Memory Limit: 65536KB
SubmitStatistic
给定一棵二叉树的先序遍历序列和中序遍历序列,要求计算该二叉树的高度。
Input
输入数据有多组,每组数据第一行输入1个正整数N(1 <= N <= 50)为树中结点总数,随后2行先后给出先序和中序遍历序列,均是长度为N的不包含重复英文字母(区分大小写)的字符串。
Output
输出一个整数,即该二叉树的高度。
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徐红博Result: AcceptedTake time: 0msTake Memory: 168KBSubmit time: 2017-02-07 15:58:38****************************************************/
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