Regex to get the size of capturing group

Update I guess you could go with some absurd Java lookahead recursion simulation that won’t work
or you could use Python to do it ?

>>> import regex
>>> rx_1_2 = r"(?m)^(1(?>(?1))*2)$"
>>>
>>> input = '''
... 111222222
... 11222234
... 1111222
... 111222
... 1122
... 12
... '''
>>> res = regex.findall( rx_1_2, input )
>>> print( res )
['111222', '1122', '12']

That this question was marked a duplicate of a Java simulated recursion
using lookaheads is astoundingly bad judgment on whoever covered this question up by marking it a duplicate. Just plain poor judgment…

It can be done with pythons regex module.
Needs to use recursion.
Done this way because it is really just nested delimiters.

1
  1
    1
    2
  2
2

1(?>[^12]++|(?R))*2

https://regex101.com/r/4Nxtvl/1

                         # Recursion 
 1                       # 1
 (?>                     # Atomic group
      [^12]++                 # Possesive, not 1 or 2
   |                        # or,
      (?R)                    # Recurse the regex
 )*                      # End cluster, do 0 to many times
 2                       # 2

To not allow inner content use 1(?>(?R))*2 https://regex101.com/r/mSUIp0/1

To add boundary conditions, contain the recursion to a group,
then surround it with boundary constructs.

(?<!\d)(1(?>[^12]++|(?1))*2)(?!\d)

https://regex101.com/r/SSr1zV/1

 (?<! \d )               # Not a digit behind
 (                       # (1 start), Recursion code group
    1                       # 1
    (?>                     # Atomic group
       [^12]++                 # Possesive, not 1 or 2
     |                        # or,
       (?1)                    # Recurse the regex group 1
    )*                      # End cluster, do 0 to many times
    2                       # 2
 )                       # (1 end)
 (?! \d )                # Not a digit ahead

To not allow inner content use (?<!\d)(1(?>(?1))*2)(?!\d) https://regex101.com/r/VI6w0Y/1