在本题中认为如果两个树左右子树交换可以相同，也被认为是同构树。

对应输入格式为：4(总结点数)

A - 1

B 2 3

C - -

D - -

#include

#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[N];

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;

}