How to change behavior of dict() for an instance

So I'm writing a class that extends a dictionary which right now uses a method "dictify" to transform itself into a dict. What I would like to do instead though is change it so that calling dict() on the object results in the same behavior, but I don't know which method to override. Is this not possible, or I am I missing something totally obvious? (And yes, I know the code below doesn't work but I hope it illustrates what I'm trying to do.)

from collections import defaultdict

class RecursiveDict(defaultdict):
    '''
    A recursive default dict.

    >>> a = RecursiveDict()
    >>> a[1][2][3] = 4
    >>> a.dictify()
    {1: {2: {3: 4}}}
    '''
    def __init__(self):
        super(RecursiveDict, self).__init__(RecursiveDict)

    def dictify(self):
        '''Get a standard dictionary of the items in the tree.'''
        return dict([(k, (v.dictify() if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

    def __dict__(self):
        '''Get a standard dictionary of the items in the tree.'''
        print [(k, v) for (k, v) in self.items()]
        return dict([(k, (dict(v) if isinstance(v, dict) else v))
                     for (k, v) in self.items()])

EDIT: To show the problem more clearly:

>>> b = RecursiveDict()
>>> b[1][2][3] = 4
>>> b
defaultdict(<class '__main__.RecursiveDict'>, {1: defaultdict(<class '__main__.RecursiveDict'>, {2: defaultdict(<class '__main__.RecursiveDict'>, {3: 4})})})
>>> dict(b)
{1: defaultdict(<class '__main__.RecursiveDict'>, {2: defaultdict(<class '__main__.RecursiveDict'>, {3: 4})})}
>>> b.dictify()
{1: {2: {3: 4}}}

I want dict(b) to be same as b.dictify()