7-3 树的同构 (25分)

7-3 树的同构 (25分)

给定两棵树T1和T2。如果T1可以通过若干次左右孩子互换就变成T2，则我们称两棵树是“同构”的。例如图1给出的两棵树就是同构的，因为我们把其中一棵树的结点A、B、G的左右孩子互换后，就得到另外一棵树。而图2就不是同构的。

图1

图2

现给定两棵树，请你判断它们是否是同构的。

输入格式:

输入给出2棵二叉树树的信息。对于每棵树，首先在一行中给出一个非负整数N (≤10)，即该树的结点数（此时假设结点从0到N−1编号）；随后N行，第i行对应编号第i个结点，给出该结点中存储的1个英文大写字母、其左孩子结点的编号、右孩子结点的编号。如果孩子结点为空，则在相应位置上给出“-”。给出的数据间用一个空格分隔。注意：题目保证每个结点中存储的字母是不同的。
输出格式:

如果两棵树是同构的，输出“Yes”，否则输出“No”。
输入样例1（对应图1）：

8
A 1 2
B 3 4
C 5 -
D - -
E 6 -
G 7 -
F - -
H - -
8
G - 4
B 7 6
F - -
A 5 1
H - -
C 0 -
D - -
E 2 -

输出样例1:

Yes

输入样例2（对应图2）：

8
B 5 7
F - -
A 0 3
C 6 -
H - -
D - -
G 4 -
E 1 -
8
D 6 -
B 5 -
E - -
H - -
C 0 2
G - 3
F - -
A 1 4

输出样例2:

No

//
// Created by TIGA_HUANG on 2020/9/24.
//

#include <iostream>
#include <sstream>

using namespace std;

struct node {
    string data;
    int left;
    int right;
    int father;
} a[20], b[20];

int s2i(string &str) {
    int ret;
    stringstream ss;
    ss << str;
    ss >> ret;
    return ret;
}

bool dfs(int t1, int t2) {
    if (t1 == -1 && t2 == -1)
        return true;
    if (t1 == -1 || t2 == -1)
        return false;
    if (a[t1].data == b[t2].data) {
        if (a[t1].left == -1 && b[t2].left == -1)
            return dfs(a[t1].right, b[t2].right);
        else if (a[t1].right == -1 && b[t2].right == -1)
            return dfs(a[t1].left, b[t2].left);
        if (a[t1].right != -1 && b[t1].right != -1 && a[a[t1].left].data == a[b[t2].left].data)
            return dfs(a[t1].left, b[t2].left) && dfs(a[t1].right, b[t2].right);
        else
            return dfs(a[t1].left, b[t2].right) && dfs(a[t1].right, b[t2].left);
    } else
        return false;
}

void checkA(int t) {
    if (t == -1)
        return;
    cout << a[t].data << ' ';
    if (a[t].left != -1)
        checkA(a[t].left);
    if (a[t].right != -1)
        checkA(a[t].right);
}

void checkB(int t) {
    if (t == -1)
        return;
    cout << b[t].data << ' ';
    if (b[t].left != -1)
        checkB(b[t].left);
    if (b[t].right != -1)
        checkB(b[t].right);
}

int main() {
    int N;
    cin >> N;
    string data, left_str, right_str;
    int left, right, root_a = -1, root_b = -1;
    for (int i = 0; i < N; ++i) {
        a[i].father = -1;
    }
    for (int i = 0; i < N; ++i) {
        cin >> data >> left_str >> right_str;
        if (left_str == "-")
            left = -1;
        else
            left = s2i(left_str);
        if (right_str == "-")
            right = -1;
        else
            right = s2i(right_str);
        a[i] = {data, left, right, a[i].father};
        if (left != -1)
            a[left].father = i;
        if (right != -1)
            a[right].father = i;
    }
    for (int i = 0; i < N; ++i) {
        if (a[i].father == -1) {
            root_a = i;
            break;
        }
    }
    cin >> N;
    for (int i = 0; i < N; ++i) {
        b[i].father = -1;
    }
    for (int i = 0; i < N; ++i) {
        cin >> data >> left_str >> right_str;
        if (left_str == "-")
            left = -1;
        else
            left = s2i(left_str);
        if (right_str == "-")
            right = -1;
        else
            right = s2i(right_str);
        b[i] = {data, left, right, b[i].father};
        b[left].father = i;
        b[right].father = i;
        if (left != -1)
            b[left].father = i;
        if (right != -1)
            b[right].father = i;
    }
    for (int i = 0; i < N; ++i) {
        if (b[i].father == -1) {
            root_b = i;
            break;
        }
    }
//    checkB(root_b);
//    cout << endl;
//    cout << root_a << ' ' << root_b;
    if (dfs(root_a, root_b))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}