数据结构与算法题目集（中文） 7-3 树的同构 (25分)

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 -

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

题解：次一个节点并不是根节点，根节点需要自己判断，可以一边建树一边判断，具体的注释在代码中。建树不难，判断那部分较难，很多人可能用的是指针形式，后面通过指针交换左右子树进行判断，哪种方法极容易错，而且麻烦，可以直接根据是否为空，不为空数值是否相等进行判断

c++：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define NULL -1
struct node
{
    char ch;
    int left;
    int right;
} tree1[100], tree2[100];
bool book[100];
//建树
int build(struct node tree[])
{
    int n;
    int book[110];
    char data, l, r;
    memset(book, 0, sizeof(book)); //将所有点初始化
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        getchar(); //吸收回车
        scanf("%c", &data);
        getchar(); //吸收空格
        scanf("%c", &l);
        getchar(); //吸收空格
        scanf("%c", &r);
        //		printf("%c %c %c\n",data,l,r);
        tree[i].ch = data;
        //判断左右子树是否存在并标记
        if (l == '-')
        {
            tree[i].left = NULL;
        }
        else
        {
            tree[i].left = l - '0';
            book[l - '0'] = 1;
        }
        if (r == '-')
        {
            tree[i].right = NULL;
        }
        else
        {
            tree[i].right = r - '0';
            book[r - '0'] = 1;
        }
    }
    int root = NULL; //这里一定要赋值为NULL
    for (int i = 0; i < n; i++)
    {
        if (!book[i])
        {
            root = i;
        }
    }
    return root;
}
bool judge(int root1, int root2)
{
    if (root1 == NULL && root2 == NULL)
    {                //判断根节点是否都为空
        return true; //都为空则相同
    }
    if (root1 == NULL && root2 != NULL || root1 != NULL && root2 == NULL)
    {
        return false; //一个根节点为空另外一个不为空则为假
    }
    if (tree1[root1].ch != tree2[root2].ch)
    {
        return false; //两棵树根节点都存在，但数值不同则为假
    }
    if (tree1[tree1[root1].left].ch == tree2[tree2[root2].left].ch)
    {
        //如果左子树的值相等则判断右子树
        return judge(tree1[root1].right, tree2[root2].right);
    }
    else
    {
        //否则判断是否第二棵树是在第一课树左右子树调换之后得到的
        return judge(tree1[root1].left, tree2[root2].right) && judge(tree1[root1].right, tree2[root2].left);
    }
}
int main()
{
    //	freopen("input.txt","r",stdin);
    int root1 = build(tree1);
    int root2 = build(tree2);
    if (judge(root1, root2))
    {
        printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }
    return 0;
}

c：

#include <stdio.h>
#include <string.h>
#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));
    char left, right;
    scanf("%d", &n);
    getchar();
    for (i = 0; i < n; i++)
    {
        scanf("%c %c %c\n", &tree[i].data, &left, &right);
        //		printf("%c %c %c\n",tree1[i].data,left,right);
        if (left == '-')
        {
            tree[i].left = NULL;
        }
        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;
    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;
    }
    if ((root1 == NULL && root2 != NULL) || (root1 != NULL && root2 == NULL))
    {
        return 0;
    }
    if (tree1[root1].data != tree2[root2].data)
    {
        return 0;
    }
    if (tree1[tree1[root1].left].data == tree2[tree2[root2].left].data)
    {
        return judge(tree1[root1].right, tree2[root2].right);
    }
    else
        return judge(tree1[root1].left, tree2[root2].right) && judge(tree1[root1].right, tree2[root2].left);
}
int main()
{
    //	freopen("input.txt","r",stdin);
    int root1, root2, i;
    root1 = build(tree1);
    root2 = build(tree2);
    //	for(i = 0;i < 8;i++)
    //	{
    //		printf("%c %d %d\n",tree1[i].data,tree1[i].left,tree1[i].right);
    //	 }
    //	 printf("\n");
    //	for(i = 0;i < 8;i++)
    //	{
    //		printf("%c %d %d\n",tree2[i].data,tree2[i].left,tree2[i].right);
    //	 }
    if (judge(root1, root2) == 1)
    {
        printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }
    return 0;
}