and definition for a Node:
write the code needed to build the structure in the above drawing
That is, given a list with n elements where n is even, [a1,a2,a3,...,an-1,an], cutInTwo truncates list to [a1,...,an/2] and returns [an/2 + 1...,a,n]. If n is odd, cutInTwo truncates list to [a1,...,an/2 + 1] and returns [an/2 + 2...,a,n].
Fill in the function below.
Node * cutInTwo(Node * list) // pre: list is NULL/0 terminated, contains n nodes // where n >= 0 // post: list contains first n/2 (rounded up) nodes, and a pointer // to the rest of the list // so that effectively the list is cut in two.