判断两棵树是否同构？

程序不正确，请大哥看看哪里出了问题？


#include<stdio.h>

#define maxsize 10

typedef struct node
{
    char element;
    int left;
    int right;
}tree;

int issame(tree firsttree[], int root1, tree sectree[], int root2);
int buildtree(tree trec[]);//纯粹建树 


int main(){
    int r1, r2, ret, i;//静态链表所构成树的根节点 
    tree t1[maxsize], t2[maxsize];//一个数组用来记录这个树 
    r1 = buildtree(t1);//构造这棵树，并返回根 
    r2 = buildtree(t2);

    if(issame(t1, r1, t2, r2))
        printf("同构\n");
    else
        printf("不同\n"); 

    return 0;
}



int buildtree(tree trec[])//用于构造一棵树，
并且把树根的位置return 
{
    int N, i,；
    char ele, l, r;
    printf("输入节点数量\n"); 
    scanf("%d", &N);
    fflush(stdin);
    int check[maxsize] = {0};//记录数组，最后遍历以找到根节点 
    if(N)
    {
        for(i=0; i<N; i++)
            check[i] = 0;    
        for(i=0; i<N; i++)
        {
            printf("请输入第%d,amount to %d times\n", i+1, N);          
            scanf("%c %d %d", &ele, &l, &r);
            fflush(stdin);

            trec[i].element = ele;//接受本节点字母 
            if(l!='-')
                {
                trec[i].left = l-'0';
                check[trec[i].left] = 1; 
                }
            else
                trec[i].left = -1; 
            if(r!='-')
                {
                trec[i].right = r-'0';
                check[trec[i].right] = 1;       
                }               
            else
                trec[i].right = -1; 
        }
    }
    for(i=0; i<N; i++)
        if(check[i] == 0)
            break;

    return i;
} 

int issame(tree firsttree[], int root1, tree sectree[], int root2)
{
    if ((firsttree[root1].element == -1)&&(sectree[root2].element == -1))//左右根节点都不存在 
        return 1;
    if(((firsttree[root1].element == -1)&&(sectree[root2].element != -1))||((firsttree[root1].element != -1)&&(sectree[root2].element == -1)))//一个有根一个没有 
        return 0;
    if(firsttree[root1].element != sectree[root2].element)//根节点元素不同 
        return 0;
    if(firsttree[root1].left == -1 && sectree[root1].left == -1) //都没有左孩子 判断右孩子 
        return (issame(firsttree, firsttree[root1].right, sectree, sectree[root2].right));
    if((firsttree[root1].left != -1)&&(sectree[root2].left != -1)&&(firsttree[firsttree[root1].left].element == sectree[sectree[root2].left].element))
    //两棵树的左孩子相同，那么就意味着不需要互换，则把左孩子和右孩子继续递归判定 
        return ((issame(firsttree, firsttree[root1].left, sectree, sectree[root2].left))&&(issame(firsttree, firsttree[root1].right, sectree, sectree[root2].right)));
    else//左右孩子相同 不需要互换 
        return (issame(firsttree, firsttree[root1].left, sectree, sectree[root2].right)&&issame(firsttree, firsttree[root1].right, sectree, sectree[root2].left)); 
}

https://blog.csdn.net/yin__ren/article/details/78992175