如何判斷兩顆二叉樹同構
#define MaxTree 10
#define ElementType char
#define Tree int
#define Null -1
struct TreeNode{
ElementType Element;
Tree Left;
Tree Right;
}T1[MaxTree],T2[MaxTree],T[MaxTree];
int main()
{
Tree R1,R2;
R1=BuildTree(T1);
R2=BuildTree(T2);
if(Isomorphic(R1,R2)) printf("Yes\n");
else printf("No\n");
return 0;
}
Tree BuildTree(struct TreeNode T[])//如何建二叉樹
{
if(N){
for(i=0;i<N;i++) check[i]=0;
for(i=0;i<N;i++){
scanf("%c %c %c\n",&T[i].Element,&cl,&cr);
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;
//對cr的對應處理
}
for(i=0;i<N;i++)
if(!check[i]) break;
Root=i;
}
return Root;
}
int Isomorphic(Tree R1,Tree R2)//如何判別兩二叉樹同構
{
if((R1==Null)&&(R2==Null))//both empty
return 1;
if((R1==Null)&&(R2==Null)||((R1!=Null)&&(R2!=Null)))
return 0;//one of them is empty
if(T1[R1].Element!=T2[R2].Element)
return 0;//roots are different
if((T1[R1].Left==Null)&&(T2[R2].Left==Null))
//both have no left subtree
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)))
//no need to swap the left and the right
return(Isomorphic(T1[R1].Left,T2[R2].Left)&&
Isomorphic(T1[R1].Right,T2[R2].Right));
else//need to swap the left and the right
return(Isomorphic(T1[R1].Left,T2[R2].Right)&&
Isomorphic(T1[R1].Right,T2[R2].Left));
}