How to completely traverse a complex dictionary of unknown depth?

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You can use a recursive generator for converting your dictionary to flat lists.

def dict_generator(indict, pre=None):
    pre = pre[:] if pre else []
    if isinstance(indict, dict):
        for key, value in indict.items():
            if isinstance(value, dict):
                for d in dict_generator(value, pre + [key]):
                    yield d
            elif isinstance(value, list) or isinstance(value, tuple):
                for v in value:
                    for d in dict_generator(v, pre + [key]):
                        yield d
            else:
                yield pre + [key, value]
    else:
        yield pre + [indict]

It returns

[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'type', u'Literal']
[u'init', u'declarations', u'body', u'value', 2]
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'i']
[u'body', u'type', u'VariableDeclaration']
[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'type', u'Literal']
[u'init', u'declarations', u'body', u'value', 4]
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'j']
[u'body', u'type', u'VariableDeclaration']
[u'body', u'kind', u'var']
[u'init', u'declarations', u'body', u'operator', u'*']
[u'right', u'init', u'declarations', u'body', u'type', u'Identifier']
[u'right', u'init', u'declarations', u'body', u'name', u'j']
[u'init', u'declarations', u'body', u'type', u'BinaryExpression']
[u'left', u'init', u'declarations', u'body', u'type', u'Identifier']
[u'left', u'init', u'declarations', u'body', u'name', u'i']
[u'declarations', u'body', u'type', u'VariableDeclarator']
[u'id', u'declarations', u'body', u'type', u'Identifier']
[u'id', u'declarations', u'body', u'name', u'answer']
[u'body', u'type', u'VariableDeclaration']
[u'type', u'Program']

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