因此,我几乎完成了这项作业,但只有在弄清楚为什么会这样做之后
当我递归调用时,此循环不会继续.

我将以名字作为输入,并返回他的孩子和父亲作为输出.

例如,Kanky的孩子是撒旦,马兰,乌兰,yugi
撒旦的孩子是本&艾伦
马兰没有孩子,我的树看起来像这样：

T=["Kanky",["satan",["ben","ian"],"Alan"],"malan",["yugi","yuppi"]]

我的代码：

def find_the_man(T,name):

    F = tree[0]     # the head of list
    C = tree[1:]    # tail of the list
    kids = ""       # find the kids and append to this str
    if F==name:
        i = 0
        while i<len(C):
          #do some things to find the children
          #return them


    for i in C:      ### this loop tries to find the man with no child
        if i==name:
            return [daddy,notfound] ### if it finds, returns it

    for i in C:             ### this is my whole thing, it takes just the
        if type(i)==list:   ### first list(unfortenately), if it finds the 
            find_the_man(i,name)  ### man by doing the actions on top,
        else:continue             ### it returns the child, but it is not able to
                              ### proceed the next lists, so when i give an input
return [notfound,notfound]    ### like (T,"yugi"), it gives notfound :(

解决方法:

老实说,我不想分析您的代码以查找问题所在,但我知道如何正确执行.

看一下这个：

def find_the_man( T, name,  p = None):
    r = False 
    for i in T: 
        if type( i ) == list: 
            r = find_the_man( i, name, T[0] )
            if r:
               break
        elif i == name:
            return ( p, [ i[0] if type(i) == list else i for i in T[ 1: ] ] ) if T.index(i) == 0 else ( T[0], None )
    return r 

T= [ "Kanky", [ "satan", [ "ben", "ian" ], "Alan" ], "malan", [ "yugi", "yuppi" ] ] 
# function return tuple ( parent, list_of_children )
find_the_man( T, "Kanky" )  # (None, ['satan', 'malan', 'yugi'])
find_the_man( T, "satan" )  # ('Kanky', ['ben', 'Alan'])
find_the_man( T, "ben" )    # ('satan', ['ian'])
find_the_man( T, "malan" )  # ('Kanky', None)
find_the_man( T, "yugi" )   # ('Kanky', ['yuppi'])
find_the_man( T, "yuppi" )  # ('yugi', None)
find_the_man(T, "stranger" )# False