I have a recursive solution to Pascal's triangle, but it returns the row of the triangle requested, and not all of the rows preceding it. I would like to know if there is a way to scoop up all the rows as they are computed from the base case and down the call stack, and return that as a list.
Writing the recursion for computing/returning any given row wasn't difficult but I thought I could just append each return to a list variable. The issue I'm having is anything I do to return the whole list messes with the return statement and breaks the computation of the row.
def pascal(n, tri):
if n == 0:
return tri
else:
r = pascal(n - 1, tri)
row = [1] + [(r[i] + r[i + 1]) for i in range(len(r) - 1)] + [1]
tri.append(row)
print('tri =', tri)
return tri[-1]
print(pascal(5, [[1]]))
The print statement inside the function shows that the rows get appended to the list. I just can't think of how to return the list outside the function. I need the last list element of 'tri' to generate the next layer, but at the same time I want to return all of 'tri' as my final return.
This is my first SO question so apologies if I'm just not seeing something blindingly obvious here. Thank you!
You should return the whole tri and use only the last element of r when you build a row:
row = [1] + [(r[-1][i] + r[-1][i + 1]) for i in range(len(r[-1]) - 1)] + [1]