Scala recursion function that removes number from list needs work -

i have recursive method in remove should zeroes given list.

def removez(list:list[int], n:int):list[int] = list match {   case nil => nil   case h::t=>     if (h == n)       t     else        h :: removez(t,n) } 

this removes 1 0 list if list has multiple zeros won't. tried adding if else statement didn't work such as:

if else(t==n)    removez(t,n) 

how can have zeroes removed?

that's because after first 0 return tail, have keep iterating:

scala> def removez(list: list[int], n: int): list[int] = list match {    |     case nil => nil    |     case h :: t =>    |       if (h == n)    |         removez(t, n) // 0 found, skip , iterate tail    |       else    |         h :: removez(t, n)    |   } removez: (list: list[int], n: int)list[int]  scala> removez(list(1,0,2,0,3), 0) res0: list[int] = list(1, 2, 3)