浙大算法与数据结构 编程题03-树1 树的同构

MOOC上有详细解释及代码陈述，在此不作赘述，注释掉代码可以看输入是否正确

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct TreeNode{
    char val;
    int Left;
    int Right;
}Tree1[105],Tree2[105];

int BuildTree(struct TreeNode Tree[],int n){
    scanf("%d",&n);
    getchar();
    char l,r;
    int index[105]={0};
    int root=-1;
    for(int i=0;i<n;i++){
        scanf("%c %c %c\n",&Tree[i].val,&l,&r);
        if(l=='-')Tree[i].Left=-1;
        else{
            Tree[i].Left=l-'0';
            index[Tree[i].Left]=1;
        }
        if(r=='-')Tree[i].Right=-1;
        else{
            Tree[i].Right=r-'0';
            index[Tree[i].Right]=1;
        }
    }
    // for(int i=0;i<n;i++)printf("%d ",index[i]);
    // printf("\n");
    for(int i=0;i<n;i++){
        if(!index[i]){
            root=i;break;
        }
    }
    return root;
}
int IfLike(int root1,int root2){
    if(root1==-1&&root2==-1)return 1;
    if(root1!=-1&&root2==-1||root1!=-1&&root2==-1)return 0;
    if(Tree1[root1].val!=Tree2[root2].val)return 0;
    if(Tree1[root1].Left==-1&&Tree2[root2].Left==-1)
        return IfLike(Tree1[root1].Right,Tree2[root2].Right);
    if(Tree1[root1].Left!=-1&&Tree2[root2].Left!=-1&&(Tree1[Tree1[root1].Left].val==Tree2[Tree2[root2].Left].val))
        return (IfLike(Tree1[root1].Left,Tree2[root2].Left)&&IfLike(Tree1[root1].Right,Tree2[root2].Right));
    else return (IfLike(Tree1[root1].Left,Tree2[root2].Right)&&IfLike(Tree1[root1].Right,Tree2[root2].Left));
}
int main(){
    int n,m;
    int root1,root2;
    root1=BuildTree(Tree1,n);
    root2=BuildTree(Tree2,m);
    // printf("root1=%d root2=%d\n",root1,root2);
    // printf("Tree1=next\n");
    // for(int i=0;i<8;i++){
    //     printf("%c %d %d\n",Tree1[i].val,Tree1[i].Left,Tree1[i].Right);
    // }
    // printf("Tree2=next\n");
    // for(int i=0;i<8;i++){
    //     printf("%c %d %d\n",Tree2[i].val,Tree2[i].Left,Tree2[i].Right);
    // }
    if(IfLike(root1,root2))printf("Yes");
    else printf("No");
}