判断数的同构

[c++]代码库

#include<bits/stdc++.h>
#define Tree int
#define ElemType char
using namespace std;
struct TreeNode
{
    ElemType element;
    Tree left;
    Tree right;
}T1[10],T2[10];
int TreeBuild(struct TreeNode T[])
{
    int N, i;
    char l, r;
    cin>> N;
    if (N)
    {
        int flag[N];
        for ( i = 0 ; i < N ; i ++)
            flag[i] = 1;
        for ( i = 0 ; i < N ; i ++)
        {
            cin>>T[i].element>>l>>r;
            if (l == '-')
                T[i].left = -1;
            else
            {
                T[i].left = l - '0';
                flag[T[i].left] = 0;
            }
            if (r == '-')
                T[i].right = -1;
            else
            {
                T[i].right = r - '0';
                flag[T[i].right] = 0;
            }
        }
        for ( i = 0 ; i < N ; i ++)
        {
            if (flag[i])
                return i;
        }
    }
    else
        return -1;
}
int judge(Tree R1, Tree R2)
{
    if ((R1 == -1) && (R2 == -1))//都是空树
        return 1;
    if ( ((R1 == -1)&&( R2 != -1)) || (R1 == -1) && ( R2 != -1))//一个是空树，一个不是空树
        return 0;
    if (T1[R1].element != T2[R2].element)//根结点元素不一样
        return 0;
    if ((T1[R1].left == -1) && (T2[R2].left == -1))//根节点都是只有右孩子，递归判断右孩子树是否同构
        return judge(T1[R1].right, T2[R2].right);
    if (((T1[R1].left != -1) && (T2[R2].left != -1)) && ((T1[T1[R1].left].element) == (T2[T2[R2].left].element)))
        return (judge(T1[R1].left, T2[R2].left) && judge(T1[R1].right, T2[R2].right));
    else
        return (judge(T1[R1].left, T2[R2].right) && judge(T1[R1].right, T2[R2].left));
}
int main ()
{
    Tree R1,R2;
    R1=TreeBuild(T1);
    R2=TreeBuild(T2);
    if(judge(R1,R2))
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
}