我想打印出一个描述乐队的简单树。我首先创建一个名为“Band”的节点,然后让孩子成为“管乐器”,然后让孩子们成为“萨克斯管”和“小号”。然后我做了一个叫做“歌曲”的“管乐器”的兄弟姐妹等等。代码非常简单:
class Node:
value = ""
down = None
right = None
def write(p):
if p==None:
return
print(p.value)
if p.down!=None: #My idea is that if we go down the tree, we indent first
print(" ",end="")
write(p.down)
write(p.right) #If we don't go down, we simply write out the siblings
a=Node()
a.value="Band"
a.down=Node()
a.down.value="Wind instruments"
a.down.down=Node()
a.down.down.value="Saxophone"
a.down.down.right=Node()
a.down.down.right.value="Trumpet"
a.down.right=Node()
a.down.right.value="Song"
a.down.right.right=Node()
a.down.right.right.value="String instruments"
a.down.right.right.down=Node()
a.down.right.right.down.value="Guitar"
a.down.right.right.down.right=Node()
a.down.right.right.down.right.value="Bass"
write(a)
输出结果为:
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass
但我希望输出为:
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass
任何人都知道如何实现这一目标?
答案 0 :(得分:8)
根据递归级别制作缩进打印,诀窍是使用一个参数来保持递归时你所处的级别:
# default with a level of 0, and an indent of 4 characters
def write(p, depth=0, indent=4):
if p==None:
return
# here we multiply the level by the number of indents
# and then you multiply that number with a space character
# which will magically show as that number of spaces.
print("{}{}".format(" "*(indent*depth), p.value))
if p.down!=None:
# then you do not need your print(…, end='') hack
# simply increase the depth
write(p.down, depth=depth+1, indent=indent)
# and for siblings, do not increase the depth
write(p.right, depth=depth, indent=indent)
我在这里使用的技巧是,默认情况下,级别为0,随着您的深入,您通过将参数增加1来增加深度。
然后,当你想要打印出缩进时,你所要做的就是将缩进字符串与该值(以及缩进大小)相乘,然后你可以根据需要打印缩进:
>>> "A"*42
'AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA'
结果是:
>>> write(a)
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass
如果你想让它更窄,因为你有很多递归:
>>> write(a, indent=1)
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass
作为奖励,我建议您将write()
功能作为Node
课程的一种方法。如果你重命名它__str__
:
class Node:
value = ""
down = None
right = None
# converts a node into a string
def as_str(self, depth=0, indent=4):
# building the current node's line, and add it to a list
ret = ["{}{}".format(" "*(indent*depth), self.value)]
if self.down:
# append down recursion first to the list
ret.append(self.down.as_str(depth=depth+1, indent=indent))
if self.right:
# then append right recursion to the list
ret.append(self.right.as_str(depth=depth, indent=indent))
# build the new string, joining each element of the list with a newline
return "\n".join(ret)
# a handler for printing the list nicely
def __str__(self):
return self.as_str()
def as_repr(self, depth=0, max_depth=2):
# building the current node's line, and add it to a list
ret = ["'{}'".format(self.value)]
if depth > max_depth:
ret.append("…")
else:
if self.down:
# append down recursion first to the list
ret.append(self.down.as_repr(depth=depth+1, max_depth=max_depth))
if self.right:
# then append right recursion to the list
ret.append(self.right.as_repr(depth=depth, max_depth=max_depth))
# build the new string, joining each element of the list with a newline
return "Node<{}>".format(",".join(ret))
# you might want to also make the repr() nicer
def __repr__(self):
return self.as_repr()
结果:
>>> a
Node<'Band',Node<'Wind instruments',Node<'Saxophone',…>,Node<'Song',Node<'String instruments',Node<'Guitar',…>>>>>
>>> print(a)
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass
HTH
答案 1 :(得分:0)
引入输入参数i
以控制要打印的缩进数。
def write(p, i = 1):
if p==None:
return
print(p.value)
j = i # store No. of current indent to j
if p.down!=None: # process p.down
for ii in range(i):
print(" ",end="")
i += 1 # No. of indent + 1
write(p.down, i)
if p.right!=None: # process p.right
for jj in range(j - 1):
print(" ",end="")
write(p.right, j) #No. of indent keep same
运行:
write(a)
结果:
Band
Wind instruments
Saxophone
Trumpet
Song
String instruments
Guitar
Bass