reverse a stack with recursion

// time: O(n^2), space: O(n)
(stack):
    for i = stack.size() -1, i > 0, i--:
        top = stack.pop()
        (stack, i, top)

// time: O(n), space: O(n)
(stack, i, top):
    if i == 0:
        stack.push(top)
        return
    // no need to check if stack empty
    temp = stack.pop()
    (stack, i -1, top)
    stack.push(temp)

OR

// time: O(n^2), space: O(n)
(stack):
    if stack.isEmpty():
        return
    temp = stack.pop()
    (stack)
    insertAtBottom(stack, temp)

// time: O(n), space: O(n)
insertAtBottom(stack, x):
    if stack.isEmpty():
        stack.push(x)
        return
    temp = stack.pop()
    insertAtBottom(stack, x)
    stack.push(temp)

[Hat tip to GfG]