实在是找不到错在哪了，怎么解决啊

Question

有人能帮忙看看，为什么这个程序过不了吗，实在是找不到错在哪了

#include<iostream>
using namespace std;

typedef enum{ATOM,LIST}ElemTag;     //ATOM==0:原子；LIST==1：子表

string emp = "()";

typedef struct GLNode
{
    ElemTag tag;
    union
    {
        char atom;
        struct { struct GLNode* hp, * tp; }ptr;
    }union1;
}*GList;

void sever(string& head, string& Tail) {
    //将非空串str分割成两部分，hstr是表头
    int n = head.size();
    int i = -1;
    int k = 0; //k记录尚未配对的“(” 数
    char ch;
    do { //搜索最外层第一个( 
        ++i;
        ch = head[i];
        if (ch == '(') 
            ++k;
        else if (ch == ')') 
            --k;

    } while (i < n && (ch != ',' || k != 0));
    if (i < n) {
        Tail = head.substr(0, i);
        //printf("in\n");
        head = head.substr(i + 1, n - i - 1);
        //printf("out\n");

    }
    else {
        Tail = head.substr(0, head.size());
        head.clear();

    }
}

void CreateGList(GList& L, string s) {
    //采用头尾链表存储结构，创建L
    if ((s.compare(emp) == 0)||s.size()==0) L = NULL;
    else {
        L = new GLNode;
        if (!L) exit(0);
        if (s.size() == 1) { //单个元素，建立原子节点 
            L->tag = ATOM;
            L->union1.atom = s[0];

        }
        else { //表节点 ,表尾 
            L->tag = LIST;
            GList p, q;
            p = L; //p是指向当前子表(表尾节点)的指针 
            string sub;
            sub = s.substr(1, s.size() - 2); //去掉外层括号 
            string hsub;
            do { //重复建立n个子表 
                sever(sub, hsub); //sub中分理出表头串hsub ，同时，sub去除了hsub
                CreateGList(p->union1.ptr.hp, hsub);
                q = p; //记录p，下面sub不为空，要再建立一个表尾节点，q记录上一层的p，用以连接q->ptr.tp = q 
                if (!sub.empty()) {
                    p = new GLNode;
                    if (!p)exit(0);
                    p->tag = LIST;
                    q->union1.ptr.tp = p;
                }
            } while (!sub.empty());
            q->union1.ptr.tp = NULL;
        }
    }
}
int Equal(GList A, GList B)
/* 判断广义表A和B是否相等,是则返回TRUE,否则返回FALSE */
{
    if (A == NULL) {
        if (B == NULL)
            return 1;
        else 
            return 0;
    }
    if (A->tag == ATOM) {
        if (B->tag == ATOM && B->union1.atom == A->union1.atom)
            return 1;
        else return 0;
    }

    if (Equal(A->union1.ptr.hp, B->union1.ptr.hp))
        if (Equal(A->union1.ptr.tp, B->union1.ptr.tp))
             return 1;
        else return 0;
    else return 0;
}


int main() {

    
    string s1;
    string s2;
    cin >> s1;
    cin >> s2;
    GList L1;
    GList L2;
    CreateGList(L1, s1);

    CreateGList(L2, s2);

    if (Equal(L1, L2))
    {
        cout << 1;
    }
    else
    {
        cout << 0;
    }

    return 0;

}

Answer 1

贴源码