如果我有动态键路径['key1'，'key2'，...，'keyN']，如何设置树的值？
from collections import defaultdict
def tree(): return defaultdict(tree)
t = tree()
def set_value(t, keys_path, value):
# TODO smth
set_value(t, ['key1', 'key2', ..., 'keyN'], 'value')
assert t['key1']['key2']...['keyN'] = 'value'
问题不是如何创建嵌套字典，就像在Nested defaultdict of defaultdict中回答的那样，而是从键列表直接访问字典的深度，因为t['key1']['key2']...['keyN']不能按原样实现