python 一个简单递归，2个小时都没写出来，牛人帮忙?

Question

import time

a = {}
a['child'] = {}
a['child']['child'] = {}
a['child']['child']['child'] = {}
print(a)


要递归给child追加{},最后输出100层

Answer 1

a = {}
for _ in range(100):
    a = {"child": a}
print(a)

Answer 2

既然要递归 写个最小白版本的

a = {}

count=0
def func(a,count):
    if count==100:
        return
    a['child']={}
    a=a['child']
    count=count+1
    func(a,count)
    

func(a,count)
print(a)

Answer 3

想到个好玩的

a = {}

exp = "a.setdefault('child', {})"
for _ in range(10):
    eval(exp)
    exp = "a['child']" + exp[1:]
print(a)

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Answer 4

def add_child(obj, n):
    if n == 0:
        return
    obj['child'] = {}
    add_child(obj['child'], n-1)

a = {}
add_child(a, 100)
print(a)

该递归函数的作用是向给定的字典obj中添加'child'键，并将其对应值设为一个新的空字典，然后在新字典中递归调用该函数。这样就可以实现向a中添加100个'child'键的操作。