Set a repeating number to 0 within a list

f1 = Flatten[Split[#] /. {x__, 1} :> {0{x} , 1}]&;
f1 @ m0

{0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1}

Also

f2 = Differences[Append[#, 0]] /. {-1 -> 1, 1 -> 0}&;
f2 @ m0

{0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1}

Update: Inspired by Mr.Wizard's post, an alternative way to use Differences:

f3 = 1 - UnitStep @ Differences @ Append[#, 0]&
f3 @ m0

{0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1}

f4 = 1 - UnitStep @ Subtract[Rest @ Append[#, 0] , #]&
f4 @ m0

{0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1}

Timings:

f5 = First /@ (Partition[Append[#, 0], 2, 1] /. {1, 1} -> {0, 0}) &; (* Fred Simons *)
f6 = Append[BitShiftRight @@@ Partition[#, 2, 1], Last @ #] &; (* march *) 
f7 = Ramp @ ListCorrelate[{1, -1}, #, 1, 0] &; (* Mr. Wizard *)
f8 = Ramp @ Differences[-Append[#, 0]] &; (* Mr. Wizard *)
f9 = FixedPoint[SequenceReplace[{1, 1} -> Sequence[0, 1]], #] &; (* Pillsy*)
f10 = Append[BitShiftRight[Most[#], Rest[#]], Last[#]] &; (*march*)

SeedRandom[1]
m0 = RandomInteger[1, 5*^5];
functions = {f1, f2, f3, f4, f5, f6, f7, f8, f9, f10};
labels = {"f1", "f2", "f3", "f4", "f5", "f6", "f7", "f8", "f9", "f10"};
{timings, results} = ConstantArray[1, {2, 10}];

Table[timings[[i]] = First[RepeatedTiming[results[[i]] = functions[[i]][m0];]], {i, 10}];

Grid[Prepend[SortBy[Transpose[{labels, functions, timings}], Last],
  {"function", SpanFromLeft, "timing"}], Dividers -> All] 

All except f9 produce the same result:

Equal @@ results

False

 Equal @@ results[[{1,2,3,4,5,6,7,8,10}]] 

True

Another one, not as nice as those of @kglr:

First /@ (Partition[Append[m0, 0], 2,1] /. {1,1}->{0,0})

(* {0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1} *)

This should work.

Append[BitShiftRight @@@ Partition[m0, 2, 1], Last@m0]

Edit: a faster version based on a comment by lasenH:

Append[BitShiftRight[Most[m0], Rest[m0]], Last[m0]]