二叉树：由已知两种遍历求未知遍历

struct node *creat(char *a, char *b, int n)
{
struct node *ptr;
char *p;
int count = 0;
if (n <= 0)
return NULL;
ptr = (struct node *)malloc(sizeof(struct node));
ptr -> data = *a;
for (p = &b[0]; p <= b + n - 1;p ++)
{
if (*p == *a)
break;
}
count = p - b;
ptr -> l = creat(a + 1, b, count);
ptr -> r = creat(a + 1 + count, p + 1, n - 1 - count);
return ptr;
}

struct node *creat(char *a, char *b, int n)
{
struct node *ptr;
if (n <= 0)
return NULL;
ptr = (struct node *)malloc(sizeof(struct node));
ptr -> data = b[n - 1];
int q = strchr(a, b[n - 1]) - a;
ptr -> l = creat(a, b, q);
ptr -> r = creat(a + q + 1, b + q, n - q - 1);
return ptr;
}