PTA 5-3 树的同构 ——理解递归
难点:
>>递归函数的实现
int Isomorphic(Tree R1,Tree R2)
{
if(R1==Null&&R2==Null)
return 1;
if(R1==Null&&R2!=Null)
return 0;
if(R1!=Null&&R2==Null)
return 0;
if(T1[R1].Element!=T2[R2].Element)
return 0;
if(T1[R1].Left==Null&&T2[R2].Left==Null)
return Isomorphic(T1[R1].Right,T2[R2].Right);
if(T1[R1].Left!=Null&&T2[R2].Left!=Null&&T1[T1[R1].Left].Element==T2[T2[R2].Left].Element)
return Isomorphic(T1[R1].Left,T2[R2].Left)&&Isomorphic(T1[R1].Right,T2[R2].Right);
else
return Isomorphic(T1[R1].Right,T2[R2].Left)&&Isomorphic(T1[R1].Left,T2[R2].Right);
}
1.由于首先在一行中给出一个非负整数N,考虑到N=0时Root=-1;
即表示根节点找不到,注意Root!=0的原因是题目默认从0开始作为下标
2.该函数传入的参数可能为Null的原因:其一,当N=0时,根节点为Null;其二,递归过程可能会出现T1[R1].Right为Null或T2[R2].Right为Null的情况;
3.当Tree R1=-1时,则该结点无意义,不能访问T1[R1].Element;
4.注意参数顺序:Tree R1延伸的结点必须在左边,Tree R2的必须在右边,应为递归中始终是T1[R1],T2[R2];
5.注意分支语句的写法
左分支相等有两种情况:一,都为Null;二,都不为Null,且值相等
请他情况则考虑交叉相等
第五点的前提是,前面有对两个都为Null,其一为Null,两者都不为Null的讨论
#include<stdio.h> #define MaxTree 11 #define ElementType char #define Tree int #define Null -1 struct TreeNode { ElementType Element;//树结点的值,字符 Tree Left; Tree Right; }T1[MaxTree],T2[MaxTree]; Tree BuildTree(struct TreeNode T[]) { int N,i,Root=-1; int check[11]; char cl,cr; scanf("%d",&N); getchar(); for(i=0;i<N;i++) check[i]=0; if(!N) return Null; if(N) { for(i=0;i<N;i++) { scanf("%c %c %c",&T[i].Element,&cl,&cr); getchar(); if(cl!='-') { T[i].Left=cl-'0'; check[T[i].Left]=1; } else T[i].Left=Null; if(cr!='-') { T[i].Right=cr-'0'; check[T[i].Right]=1; } else T[i].Right=Null; } } for(i=0;i<N;i++) if(!check[i]) return i; } int Isomorphic(Tree R1,Tree R2) { if(R1==Null&&R2==Null) return 1; if(R1==Null&&R2!=Null) return 0; if(R1!=Null&&R2==Null) return 0; if(T1[R1].Element!=T2[R2].Element) return 0; if(T1[R1].Left==Null&&T2[R2].Left==Null) return Isomorphic(T1[R1].Right,T2[R2].Right); if(T1[R1].Left!=Null&&T2[R2].Left!=Null&&T1[T1[R1].Left].Element==T2[T2[R2].Left].Element) return Isomorphic(T1[R1].Left,T2[R2].Left)&&Isomorphic(T1[R1].Right,T2[R2].Right); else return Isomorphic(T1[R1].Right,T2[R2].Left)&&Isomorphic(T1[R1].Left,T2[R2].Right); } int main() { Tree R1,R2; R1=BuildTree(T1); R2=BuildTree(T2); if(Isomorphic(R1,R2)) printf("Yes\n"); else printf("No\n"); return 0; }