原来的代码没处理dict为空的情况
1 def flatten(dictionary):
2 #[] is a list
3 #() is a tuple
4 stack = [((), dictionary)]
5
6 result = {} #result is a dict
7
8 while stack:
9 path, current = stack.pop() #get a tuple
10
11 for k, v in current.items(): #dict::items return key and values tuple
12 if isinstance(v, dict): #is a instance of dict
13 stack.append((path + (k,), v)) #add key to tuple such as (xxx, yyy, zzz) and the element in stack is like ((xxx, yyy, zzz), value)
14 else:
15 result["/".join((path + (k,)))] = v
16
17 if len(current) == 0: #when the dict is empty
18 result["/".join(path)] = ""
19
20 return result
(k,)一个元素的tuple