7-3 树的同构 (25分)

#define null -1

struct node {
    int left, right;
    char data;
} tree1[100], tree2[100];

int book[1000];

int build(struct node tree[]) {
    int n, i, root;
    memset(book, 0, sizeof(book));  //节点初始化
    char left, right;
    scanf("%d", &n);
    getchar();
    for (i = 0; i < n; i++) {
        scanf("%c %c %c\n", &tree[i].data, &left, &right);
        if (left == '-') tree[i].left = null; //判断左右子树是否存在并标记
        else {
            tree[i].left = left - '0';
            book[left - '0'] = 1;
        }
        if (right == '-') tree[i].right = null;
        else {
            tree[i].right = right - '0';
            book[right - '0'] = 1;
        }
    }
    root = null; //必须初始化
    for (i = 0; i < n; i++) {
        if (book[i] == 0) root = i;
    }
    return root;
}

int judge(int root1, int root2) {
    if (root1 == null && root2 == null) return 1; // 根节点都为空则相同
    if ((root1 == null && root2 != null) || (root1 != null && root2 == null)) return 0; //一个空一个不空
    if (tree1[root1].data != tree2[root2].data) return 0; //两个根节点相同，但数值不同
    if (tree1[tree1[root1].left].data == tree2[tree2[root2].left].data) { //左子树的值相等，则判断右子树
        return judge(tree1[root1].right, tree2[root2].right);
    }
    else { //调换第一棵树看是否能得到第二棵树
        return judge(tree1[root1].left, tree2[root2].right) && judge(tree1[root1].right, tree2[root2].left);
    }
}

int main() {
    int root1, root2;
    root1 = build(tree1);
    root2 = build(tree2);
    judge(root1, root2) == 1 ? printf("Yes\n") : printf("No\n");
    return 0;
}