Question

来自http://cbio.ufs.ac.za/live_docs/nbn_tut/trees.html

让我们创建一个python类来表示一棵树。我们需要一些方法将数据存储在节点中，并以某种方式指示任何子节点，或子树。
class node(object):
    def __init__(self, value, children = []):
        self.value = value
        self.children = children
哇！这似乎太容易了......不管你信不信，它确实如此工作。让我们使用我们的新课来存储我们的家谱...
tree = node("grandmother", [
    node("daughter", [
        node("granddaughter"),
        node("grandson")]),
    node("son", [
        node("granddaughter"),
        node("grandson")])
    ]);

我希望能够同时获取每个node实例的子项和父项，所以我认为我需要定义其父项及其子项

class node(object):
    def __init__(self, value, children = [], parent = []):
        self.value = value
        self.children = children
        self.parent = parent

但问题是每个节点的子节点和父节点都会有一个副本。如果我更改其值，我将不得不更改其副本中的所有值。在C ++中，没有这样的问题，因为我们可以通过将指针存储到其子节点以及仅存在于其中的父节点来引用节点的子节点和父节点。我想知道如何在Python中实现这样的树？感谢。

Answer 1

您可以在节点构造函数中指定子项的父项：

class node(object):
    def __init__(self, value, children = None):
        self.value = value
        self.children = children or []
        self.parent = None
        for child in self.children:
            child.parent = self

Answer 2

class node(object): 
  def __init__(self, value, children = []): 
    self.value = value 
    self.children = children

tree = [node("grandmother", [ node("daughter", [ node("granddaughter"), node("grandson")]), node("son", [ node("granddaughter"), node("grandson")]) ])];

def familyValues(targetName, siblings = []):
  family = []
  for sibling in siblings:
    if sibling.value == targetName:
      family.append(sibling)
      family = family + sibling.children
      break
    else:
      children = familyValues(targetName, sibling.children)
      if len(children) > 0:
        children.append(sibling)
        family = children

  return family

myFamily = familyValues('daughter', tree)
for sibling in myFamily:
  print(sibling.value)

除非你明确克隆它们，否则

没有python中的对象副本

实现一个树，孩子和父母可以互相参考

2 个答案: