# The (Easy) Road to Code

## Python 2, 797873 72 bytes

n,d=input()
c='|\/'[d]
i=n
while i:print' '*(n-i*d)+c,i%2*c or' ',c;i-=1


Try it online!

Takes [1,0,-1] for [north-west, north, north-east]

-1 byte, thanks to Neil

## Python 2, 66 bytes

n,d=input()
s=' '+'|\/'[d]
for c in(s*n)[n:]:print' '*n+s,c+s;n+=d


Try it online!

Uses d=-1 for NE, d=0 for N, and d=1 for NW. Takes advantage of leading spaces being allowed. The rule that the bottom segment of the road have a separator made it tricky to get the parity right; it's achieved by the slicing (s*n)[n:] taking the second half of the 2n alternations between space and a the road character.

## 1. Python 3.5, 122 120 bytes

Script takes two params: n, d.

d: 0, 1, 2 -> \ | /

tio.run

import sys;(n,d)=[*map(int,sys.argv[1:3])];c="\\|/"[d]
for i in range(n):j=n+~i;print(" "*(i,0,j)[d],c,c*(j%2<1)or" ",c)


output:

$./script.py 6 2 / / / / / / / / / / / / / / /$ ./script.py 6 1
|   |
| | |
|   |
| | |
|   |
| | |
\$ ./script.py 6 0
\   \
\ \ \
\   \
\ \ \
\   \
\ \ \


### Explanation

# parse input params
(n,d)=[*map(int,sys.argv[1:3])]

c="\\|/"[d]

# loop n-times
for i in range(n):

# complement (how many lines to end)
j=n+~i

# print
#   space i or 0 or j times