03-树1 树的同构

问题

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

解答

#include <iostream>

using namespace std;

template <typename T>
class Node
{
public:
    T data;
    int left;
    int right;
};

template <typename T>
class Tree
{
public:
    unsigned N;
    Node<T> *tree;
    int root;
    Tree(int n) : N(n)
    {
        if (n == 0)
        {
            N = 0;
            tree = nullptr;
            root = -1;
            return;
        }
        root = 0;
        tree = new Node<T>[N];
        char c, l, r;
        for (int i = 0; i < n; ++i)
        {
            cin >> c >> l >> r;
            tree[i].data = c;
            tree[i].left = l == '-' ? -1 : l - '0';
            tree[i].right = r == '-' ? -1 : r - '0';
            if (tree[i].left != -1)
                root ^= tree[i].left;
            if (tree[i].right != -1)
                root ^= tree[i].right;
        }
    }
    ~Tree()
    {
        if (N)
            delete[] tree;
    }
};

bool isomorphism(int r1, int r2, const Tree<char> &t1, const Tree<char> &t2);

int main()
{
    int n = 0;
    cin >> n;
    Tree<char> t1(n);
    cin >> n;
    Tree<char> t2(n);
    if (isomorphism(t1.root, t2.root, t1, t2))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

bool isomorphism(int r1, int r2, const Tree<char> &t1, const Tree<char> &t2)
{
    if (r1 == -1 and r2 == -1)
        return true;
    if ((r1 == -1 and r2 != -1) or (r1 != -1 and r2 == -1))
        return false;
    if (t1.tree[r1].data != t2.tree[r2].data)
        return false;
    if (t1.tree[r1].left == -1 and t2.tree[r2].left == -1)
        return isomorphism(t1.tree[r1].right, t2.tree[r2].right, t1, t2);
    if ((t1.tree[r1].left != -1 and t2.tree[r2].left != -1) and (t1.tree[t1.tree[r1].left].data == t2.tree[t2.tree[r2].left].data))
        return (isomorphism(t1.tree[r1].left, t2.tree[r2].left, t1, t2) and isomorphism(t1.tree[r1].right, t2.tree[r2].right, t1, t2));
    else
        return (isomorphism(t1.tree[r1].left, t2.tree[r2].right, t1, t2) and isomorphism(t1.tree[r1].right, t2.tree[r2].left, t1, t2));
}