[TheLoneWolfling]

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.