Iterative solution:

    def invertTree(self, root):
        queue = []
        if root is not None:
            queue.append(root)
        while queue:
            current = queue.pop()
            current.left, current.right = current.right, current.left
            if current.left is not None:
                queue.append(current.left)
            if current.right is not None:
                queue.append(current.right)
        return root

Better solution:

    def invertTree(self, root):
        queue = []
        if root is not None:
            queue.append(root)
        while current is not None or len(queue) > 0:
            while current is None:
                current = queue.pop()
            current.left, current.right = current.right, current.left
            if current.left is not None:
                if current.right is not None:
                    queue.append(current.right)
                current = current.left
            else:
                current = current.right
        return root

No needless enqueuing / dequeuing of things unless you need to. As a bonus, if the nodes maintain a count of how many children the have overall you can always recurse on the smaller child first and use less memory.

Any particular reason why you're using an array as a stack (LIFO) but calling it "queue" (FIFO)?

Not sure why, but I'd feel iffy iterating on a list while mutating its content.

But I guess it's file while ignoring the order of the items, and just caring if it's containing items or empty.

Carry on :)

A functional approach:

      (defun mirror (tree)
        (when tree
          (tree 
              (tree-data tree)
              (mirror (tree-right tree))
              (mirror (tree-left tree)))))

For completness, here are the definitions:

      (defpackage :trees (:use :cl)
                  (:shadow #:copy-tree))
      (in-package :trees)
      (defstruct (tree (:constructor tree (data &optional left right))
                       (:type list))
        data left right)

As well as a test case:

      (mirror (tree 4 
                    (tree 2
                          (tree 1)
                          (tree 3))
                    (tree 7
                          (tree 6)
                          (tree 9))))

      => (4 (7 (9 NIL NIL) (6 NIL NIL)) (2 (3 NIL NIL) (1 NIL NIL)))

For everyone who enjoyed this or wants to take it a step further, I recommend writing a CHIP-8 emulator. I used the following source: http://www.multigesture.net/articles/how-to-write-an-emulato... and it was very helpful.

Bravo to the CEO for handling it this way. It really shows how treating others like humans and remaining compassionate no matter what the situation is beneficial to everyone.