Recursion: Reversed List Index

Write a recursive function that filters the items in a list (given as parameter arr) by positional parity (odd or even), given as parameter par, starting from the opposite end. Return a list of items on odd positions (... 5, 3, 1) or on even positions (... 6, 4, 2) and counting from the last item in the list.

Examples

get_items_at([2, 4, 6, 8, 10], "odd") ➞ [2, 6, 10]
// 2, 6 & 10 occupy the 5th, 3rd and 1st positions from right.
// Odd positions, hence the parity, and from the opposite.

get_items_at(["E", "D", "A", "B", "I", "T"], "even") ➞ ["E", "A", "I"]
// E, A and I occupy the 6th, 4th and 2nd positions from right.
// Even positions, hence the parity, and from the opposite.

get_items_at([")", "(", "*", "&", "^", "%", "$", "#", "@", "!"], "even") ➞ [")", "*", ^", "$", "@"]

get_items_at(["A", "R", "B", "I", "T", "R", "A", "R", "I", "L", "Y"], "even") ➞ ["R", "I", "R", "R", "L"]

Notes

• IMPORTANT: You are advised to solve this challenge via a recursive approach, hence, the very purpose of this challenge.
• Lists are zero-indexed, so, index+1 = position or position-1 = index.
• Items in the list may contain duplicates. See example #4.
• The last item in the list is always the first item to start a positional count.
• The sequence of the items in the resulting list should be retained (i.e. example #1 - 6 should come after 2 and before 10, example #2 - "A" should come after "E" and before "I").
• An iterative version of this challenge can be found via this link.