c语言实现判断两颗树是否同构

#include <stdio.h>
#define Tree int
#define Null -1
#define MAXSIZE 10

struct Node{
char Element;
Tree Left;
Tree Right;
}T1[MAXSIZE], T2[MAXSIZE];

Tree BuildTree(struct Node T[])
{
int N, i, Root;
char ch, cl, cr;

scanf("%d", &N);
ch = getchar();

int Check
;

for(i = 0; i < N; i++) Check[i] = 0;	//为了标记根节点 ，因为没有节点指向根节点
for(i = 0; i < N; i++){
scanf("%c %c %c", &T[i].Element, &cl, &cr);
ch = 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++)			//遍历找到根节点
if(Check[i] == 0)
break;
Root = i;
return Root;
}

int Isomorphic(Tree R1, Tree R2)
{
//两树都为空
if(R1 == Null && R2 == Null)
return 1;
//空树和非空树
if((R1 == Null && R2 != Null) || (R1 != Null && R2 == Null))
return 0;
//两树的根节点不一样
if(T1[R1].Element != T2[R2].Element)
return 0;
//若过两树的左子树都空，判断右子树是否一样
if(T1[R1].Left == Null && T2[R2].Left == Null)
return Isomorphic(T1[R1].Right, T2[R2].Right);
//若两树的左子树不空，并且左子树的结点元素都一样，判断左右子树是否一样
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
//否则 判断两树是否同构
return Isomorphic(T1[R1].Left, T2[R2].Right) && Isomorphic(T1[R1].Right, T2[R2].Left);
}

int main()
{
Tree R1, R2;

R1 = BuildTree(T1);
R2 = BuildTree(T2);
if(Isomorphic(R1, R2)) printf("Yes");
else
printf("No");
return 0;
}