基础实验4-2.1 树的同构 (25 分)

判断逻辑和分析：

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxv = 20;
struct node
{
	char data;
	int left,right;
};
node T1[maxv],T2[maxv];
bool compare(int r1, int r2)
{
	if(r1==-1&&r2==-1) return true;
	if((r1==-1&&r2!=-1)||(r1!=-1&&r2==-1)) return false;
	if(T1[r1].data!=T2[r2].data) return false;
	/*下面就是默认根存在且相等*/
	if(T1[r1].left==-1&&T2[r2].left==-1)
		return compare(T[r1].right,T[r2].right);
	if(T1[r1].left!=-1&&T2[r2].left!=-1&&T1[T1[r1].left].data==T2[T2[r2].left].data)
		return (compare(T1[r1].left,T2[r2].left)&&compare(T1[r1].right,T2[r2].right));
	else
		return (compare(T1[r1].left,T2[r2].right)&&compare(T1[r1].right,T2[r2].left));
}
int main()
{
	int n1,n2,i;
	bool A[maxv],B[maxv];
	fill(A,A+maxv,false); fill(B,B+maxv,false);
	char data,left,right,bug;
	scanf("%d\n",&n1);
	for(i=0; i<n1; i++)
	{	
		scanf("%c %c %c\n",&data,&left,&right);
		T1[i].data = data;
		if(left>='0'&&left<='9') { T1[i].left = left - '0'; A[T1[i].left] = true;}
		else T1[i].left = -1;
		if(right>='0'&&left<='9') { T1[i].right = right - '0'; A[T1[i].right] = true;}
		else T1[i].right = -1;
	}
	scanf("%d\n",&n2);
	for(i=0; i<n2; i++)
	{	
		scanf("%c %c %c\n",&data,&left,&right);
		T2[i].data = data;
		if(left>='0'&&left<='9') { T2[i].left = left - '0'; B[T2[i].left] = true;}
		else T2[i].left = -1;
		if(right>='0'&&left<='9') { T2[i].right = right - '0'; B[T2[i].right] = true;}
		else T2[i].right = -1;
	}
	if(n1==0&&n2==0) {printf("Yes\n"); return 0;}
	if(n1!=n2) { printf("No\n"); return 0; }
	for(i=0; i<n1; i++)
	{
		if(A[i]==false) break;
	}
	int r1 = i;
	for(i=0; i<n2; i++)
	{
		if(B[i]==false) break;
	}
	int r2 = i;
	bool ans = compare(r1,r2);
	if(ans==true) printf("Yes\n");
	else
		printf("No\n");
	return 0;
}