POJ 2255 Tree Recovery [二叉树]

来源：互联网发布：sinead o connor 知乎编辑：程序博客网时间：2024/05/29 15:10

题意：

给前序和中序，求后序便利。

思路：

惭愧，把最原始的都忘了，记得这应该是数据结构的例题之类的知识。

所以特地把最原始的树也构造一遍，再后序输出，练练手。

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<vector>#include<queue>#include<cmath>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))#define Abs(a) ((a)>0?(a):(-(a)))#define llong long long intusing namespace std;const int N=30;const int inf=(1<<30);int n,m;char str1[N],str2[N];struct Node{int father;int leftson;int rightson;}node[N];void preorder(char * s1,char * s2,int len,int pre,int son)//rebuild the tree{int now=s1[0]-'A'+1;if(len==1){if(son)node[pre].rightson=now;else  node[pre].leftson=now;node[now].father=pre;return;}int pos;for(pos=0;pos<len;pos++){if(s2[pos]==s1[0]){break;}}if(pos){preorder(s1+1,s2,pos,now,0);}if(len-pos-1){preorder(s1+pos+1,s2+pos+1,len-pos-1,now,1);}if(son)  node[pre].rightson=now;else  node[pre].leftson=now;node[now].father=pre;}void print_pod(int now){if(!now)  return;print_pod(node[now].leftson);print_pod(node[now].rightson);printf("%c",now+'A'-1);}int main(){while(scanf(" %s %s",str1,str2)!=EOF){memset(node,0,sizeof(node));n=strlen(str1);preorder(str1,str2,n,0,0);print_pod(str1[0]-'A'+1);printf("\n");}return 0;}