S > -SS
S > 0
S > 1

S > SS-
S > 0
S > 1

Prefix notation:  --01-0-01
Parentheses:      -(-01)(-0(-01))
Convert:          (01-)(0(01-)-)-
Postfix notation: 01-001---

Input       Output
1           1
0           0
-01         01-
-10         10-
--01-0-01   01-001---

,[[->++++<<+>]>[[-]<<[.[-]<]]>,]

,[                Take input and start main loop
  [->++++<<+>]    Push input, and compute 4*input
  >[              If 4*input is nonzero (and thus input is not @):
    [-]<<           Zero out this cell and move to top of stack
    [.[-]<]         Pop from stack and output until  is reached
  ]
  >,              Move pointer into the correct position.  If input was @, the earlier > pushed  onto the stack.
]

M!`-*.
+m`^-(.*)¶(d.*)
$1$2-

def p(s):x=next(s);yield from[x]*(x>"-")or[*p(s),*p(s),"-"]

>>> list(p(iter("--01-0-01")))
['0', '1', '-', '0', '0', '1', '-', '-', '-']

L+&-hbTsyM.-Btbytbhb

L+&-hbTsyM.-Btbytbhb
L                      define a function y(b), that returns:
   -hbT                   remove the chars "10" from the first char b
                          (T=10, and - will convert a number to a string)
  &                       if this gives the empty string (a falsy value)
 +                hb         then append b[0] to it and return it
                             (so this will parse a digit 0 or 1 from the string)
  &                       otherwise (the first char is a -)
               ytb           parse the first prefix expression from b[1:]
                             (recursive call)
          .-Btb              remove this parsed expression bifurcated from b[1:]
                             this gives a tuple [b[1:], b[1:] without first expr]
        yM                   parse and convert an expression from each one
       s                     join the results
 +                hb         and append the b[0] (the minus) to it and return

;
-;
¶
+`¶(.+);(.+)
$1$2-
;