Symmetric Cases

Everything we did for the front exists for the back, with head and tail swapped and next and prev swapped. The _mut peek variants are the same shape but call borrow_mut and return RefMut<T> instead of Ref<T>. Boilerplate, no new ideas.

push_back, pop_back, peek_back: text-substitute the front versions. head becomes tail, tail becomes head, next becomes prev, prev becomes next. Same code, mirrored.

pub fn push_back(&mut self, elem: T) {
    let new_tail = Rc::new(RefCell::new(Node { elem, prev: None, next: None }));
    match self.tail.take() {
        Some(old_tail) => {
            old_tail.borrow_mut().next = Some(new_tail.clone());
            new_tail.borrow_mut().prev = Some(old_tail);
            self.tail = Some(new_tail);
        }
        None => {
            self.head = Some(new_tail.clone());
            self.tail = Some(new_tail);
        }
    }
}

pub fn pop_back(&mut self) -> Option<T> {
    self.tail.take().map(|old_tail| {
        match old_tail.borrow_mut().prev.take() {
            Some(new_tail) => {
                new_tail.borrow_mut().next.take();
                self.tail = Some(new_tail);
            }
            None => {
                self.head.take();
            }
        }
        Rc::try_unwrap(old_tail).ok().unwrap().into_inner().elem
    })
}

The _mut peek variants exist because mutable access through a RefCell is a different guard type. borrow_mut() returns RefMut<'_, T>, which is exclusive. RefMut::map projects it the same way Ref::map does for shared.

use std::cell::RefMut;

pub fn peek_back(&self) -> Option<Ref<T>> {
    self.tail.as_ref().map(|n| Ref::map(n.borrow(), |n| &n.elem))
}

pub fn peek_front_mut(&mut self) -> Option<RefMut<T>> {
    self.head.as_ref().map(|n| RefMut::map(n.borrow_mut(), |n| &mut n.elem))
}

pub fn peek_back_mut(&mut self) -> Option<RefMut<T>> {
    self.tail.as_ref().map(|n| RefMut::map(n.borrow_mut(), |n| &mut n.elem))
}