I found an interesting problem on https://leetcode.com/problems/sequential-digits/:
An integer has sequential digits if and only if each digit in the number is one more than the previous digit.
Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits.
 
Example 1:
Input: low = 100, high = 300
Output: [123,234]
Example 2:
Input: low = 1000, high = 13000
Output: [1234,2345,3456,4567,5678,6789,12345]

 
Constraints:

10 <= low <= high <= 10^9

I tried it in Python, but could not do it. Can you solve it?
class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        nums = []
        max_digits = 9
        for depth in range(2, max_digits + 1):
            nums += self.genNum(max_digits, depth)
        return [n for n in nums if n >= low and n <= high]
    
    def genNum(self, max_digits, depth):
        result = []
        for startDigit in range(1, max_digits-depth+2):
            num = startDigit
            for lvl in range(depth-1):
                num = num * 10 + num % 10 + 1
            result.append(num)
        return result