以下是对链表按字母顺序和频率顺序进行排序的一个解决方法的示例代码：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def sortLinkedList(head):
    if not head or not head.next:
        return head

    # 使用字典统计每个字母的频率
    freq = {}
    current = head
    while current:
        if current.val not in freq:
            freq[current.val] = 1
        else:
            freq[current.val] += 1
        current = current.next

    # 根据字母顺序和频率顺序重新构建链表
    sorted_chars = sorted(freq.keys())
    dummy = ListNode(0)
    current = dummy
    for char in sorted_chars:
        count = freq[char]
        while count > 0:
            current.next = ListNode(char)
            current = current.next
            count -= 1

    return dummy.next

使用示例：

# 创建一个示例链表
head = ListNode('a')
head.next = ListNode('b')
head.next.next = ListNode('b')
head.next.next.next = ListNode('c')
head.next.next.next.next = ListNode('a')
head.next.next.next.next.next = ListNode('a')

# 对链表进行排序
sorted_head = sortLinkedList(head)

# 打印排序后的链表
current = sorted_head
while current:
    print(current.val, end=' ')
    current = current.next

# 输出: a a a b b c

以上代码首先遍历链表，统计每个字母的频率，并将频率信息存储在一个字典中。然后，根据字母顺序和频率顺序重新构建链表。最后，打印排序后的链表的值。