Question

假设我有这棵树：

                                     cough
                      Yes /                            \ No
                   sneezing                        sneezing
               Yes /       \ No                Yes /       \ No
             fever         fever               fever         fever
       Yes /    \ No    Yes/     \No       Yes /    \ No    Yes/     \No
       dead   cold   influenza   cold      dead   influenza cold   healthy

我想要疾病的途径"influenza" 输出应该是这样的：

[[True,False,True],[False,True,False]]

如果你走到根的右边，它会返回True（是），如果你去左边是假的（不）

这是我一直试图为这个函数做的代码但是我做错了它不会像我想要的那样返回..

def paths_to_illness(self, illness):

    head=self.__root
    new_list=[]
    new_list=diagnoser.get_path(head,illness)
    return new_list

def get_path(self,head,illness):

    if head is None:
        return []
    if (head.positive_child == None and head.negative_child==None):
        return [head.data]

    left_tree=diagnoser.get_path(head.negative_child,illness)
    right_tree=diagnoser.get_path(head.positive_child,illness)
    all_tree=left_tree+right_tree

    list1=[]

    for leaf in  all_tree:

        if illness == leaf:

            list1.append(["True"])
        else:
            list1.append(["False"])

    return list1

有什么想法可以帮助我修复代码吗？感谢

diagonser只是一个不重要的类，我的节点类右侧为"positive_child"而左"negative_child"

如果还有其他不清楚的地方请告诉我

谢谢！

根据要求提出制作树的代码：

class Diagnoser:
def __init__(self, root):
    self.__root = root


class Node:
 def __init__(self, data="", pos=None, neg=None):
    self.data = data
    self.positive_child = pos
    self.negative_child = neg


leaf1 = Node("dead", None, None)
leaf2 = Node("cold", None, None)
fever1 = Node("fever", leaf1, leaf2)

leaf3 = Node("influenza", None, None)
leaf4 = Node("cold", None, None)
fever2 = Node("fever", leaf3, leaf4)

sneezing1 = Node("sneezing", fever1, fever2)

leaf5 = Node("dead", None, None)
leaf6 = Node("influenza", None, None)
fever3 = Node("fever", leaf5, leaf6)

leaf7 = Node("cold", None, None)
leaf8 = Node("healthy", None, None)
fever4 = Node("fever", leaf7, leaf8)

sneezing2 = Node("sneezing", fever3, fever4)

root = Node("cough", sneezing1, sneezing2)
diagnoser = Diagnoser(root)

Answer 1

这就是我提出的

class Tree:
  def __init__(self, data, left=None, right=None):
    self.data = data
    self.left = left
    self.right = right

  @property
  def is_leaf(self):
    return not (self.left or self.right)

  def __repr__(self):
    return 'Tree({}, {}, {})'.format(self.data, self.left, self.right)

  def find(self, target, path_to=()):
    if self.is_leaf:
      if self.data == target:
        yield path_to
    else:
      if self.left:
        yield from self.left.find(target, (*path_to, True))
      if self.right:
        yield from self.right.find(target, (*path_to, False))

t = Tree('Cough', Tree('Sneezing', Tree('Fever', Tree('Dead'), Tree('Cold')), Tree('Fever', Tree('Influenza'), Tree('Cold'))), Tree('Sneezing', Tree('Fever', Tree('Dead'), Tree('Influenza')), Tree('Fever', Tree('Cold'), Tree('Healthy'))))

print(list(t.find('Influenza')))

通过让我们的find方法成为生成器，我们可以使用yield from轻松地在调用堆栈中冒出正面结果。如果您使用的是不支持参数解包(*path_to, True)的Python版本，则path_to + (True,)是等效的

编辑：这是一个不使用yield

的版本

def find(self, target, path_to=()):
  if self.is_leaf:
    if self.data == target:
      return [path_to]
    else:
      return []
  else:
    if self.left:
      l = self.left.find(target, (*path_to, True))
    if self.right:
      r = self.right.find(target, (*path_to, False))
    return l + r

Python检查叶子的路径在二进制树python给叶子的数据

1 个答案: