03-树1 树的同构

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。


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NN (\le 10≤10)，即该树的结点数（此时假设结点从0到N-1N−1编号）；随后NN行，第ii行对应编号第ii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

题目描述：一棵树左右孩子对调，为同构，若不是左右孩子对调，则不是，现在判断一棵树是否同构。

思路：采用递归来判断。
1.AB都空，y
2.AB其中一个不空，n
3.AB的数据不相等，n
4.A树的左孩子B树的左孩子空，return 判断A树的右孩子和B树的右孩子
5.一个空一个不空，return A左与B右&&A右与B左
6.A树的左孩子B树的左孩子不空，return A左与B左&&A右与B右

#include <iostream>using namespace std;#define Elementtype chartypedef struct bitnode{    Elementtype data;    char left;    char right;}bitnode;int findtop(int *findtop, int num){    for (int i = 0; i < num; i++)    {        if (findtop[i] != 1)            return i;    }}bitnode* tree_1;bitnode* tree_2;int travel(int t1, int t2){    if (t1 == -1 && t2 == -1)        return 1;    if ((t1 != -1 && t2 == -1) || (t1 == -1 && t2 != -1))        return 0;    if (tree_1[t1].data != tree_2[t2].data)        return 0;    if ((tree_1[t1].left == -1) && (tree_2[t2].left == -1))        return travel(tree_1[t1].right, tree_2[t2].right);    if (tree_1[tree_1[t1].left].data == tree_2[tree_2[t2].left].data)        return (travel(tree_1[t1].left, tree_2[t2].left) && travel(tree_1[t1].right, tree_2[t2].right));    else        return (travel(tree_1[t1].left, tree_2[t2].right) && travel(tree_1[t1].right, tree_2[t2].left));}void input(){    int treenum_1 = 0, treenum_2 = 0;    cin >> treenum_1;    tree_1 = new bitnode[treenum_1];    int *findtree_1top = new int[treenum_1];    char temp1;    char temp2;    for (int i = 0; i < treenum_1; i++)    {        cin >> (tree_1 + i)->data;        cin >> temp1;        cin >> temp2;        if (temp1 == '-')            (tree_1 + i)->left = -1;        else        {            (tree_1 + i)->left = (temp1 - '0');            findtree_1top[temp1 - '0'] = 1;            //cout << (int)(tree_1 + i)->left << ' ';        }        if (temp2 == '-')            (tree_1 + i)->right = -1;        else        {            (tree_1 + i)->right = (temp2 - '0');            findtree_1top[temp2 - '0'] = 1;            //cout << (int)(tree_1 + i)->right << ' ';        }    }    cin >> treenum_2;    int *findtree_2top = new int[treenum_2];    if (treenum_1 == treenum_2)    {        if (treenum_1 == 0)        {            cout << "Yes";        }        else        {            tree_2 = new bitnode[treenum_2];            for (int i = 0; i < treenum_2; i++)            {                cin >> (tree_2 + i)->data;                cin >> temp1;                cin >> temp2;                if (temp1 == '-')                    (tree_2 + i)->left = -1;                else                {                    (tree_2 + i)->left = (temp1 - '0');                    findtree_2top[temp1 - '0'] = 1;                    //cout << temp1 << endl;                }                if (temp2 == '-')                    (tree_2 + i)->right = -1;                else                {                    (tree_2 + i)->right = (temp2 - '0');                    findtree_2top[temp2 - '0'] = 1;                    //cout << temp2 << endl;                }            }            int treetop1 = findtop(findtree_1top, treenum_1);            int treetop2 = findtop(findtree_2top, treenum_2);            if (travel(treetop1, treetop2))                cout << "Yes";            else                cout << "No";        }    }    else    {        cout << "No";    }    delete[]findtree_1top;    delete[]findtree_2top;    delete[]tree_1;    delete[]tree_2;}int main(){    input();    return 0;}