Python 如何从特定节点获取树祖先的列表?
给定一棵树,其中每个节点可以有N个子节点,但只有一个父节点。如何获取一个节点的祖先?例如,假设我得到了这棵树:Python 如何从特定节点获取树祖先的列表?,python,python-3.x,tree,ancestor,Python,Python 3.x,Tree,Ancestor,给定一棵树,其中每个节点可以有N个子节点,但只有一个父节点。如何获取一个节点的祖先?例如,假设我得到了这棵树: # Operator # ... FooOperator # ...... BOperator # ......... B1Operator # ............ B11Operator # ...... AOperator # ......... A2Operator # ......... A1Operator # ......... A3Operator # .....
# Operator
# ... FooOperator
# ...... BOperator
# ......... B1Operator
# ............ B11Operator
# ...... AOperator
# ......... A2Operator
# ......... A1Operator
# ......... A3Operator
# ...... COperator
# ......... C1Operator
# ......... C2Operator
# ............ C21Operator
tree = {
'children': [{
'children': [{
'children': [{
'children': [{
'children': [],
'class': 'B11Operator',
'parent': 'B1Operator'
}],
'class': 'B1Operator',
'parent': 'BOperator'
}],
'class': 'BOperator',
'parent': 'FooOperator'
},{
'children': [{
'children': [],
'class': 'A2Operator',
'parent': 'AOperator'
},{
'children': [],
'class': 'A1Operator',
'parent': 'AOperator'
},{
'children': [],
'class': 'A3Operator',
'parent': 'AOperator'
}],
'class': 'AOperator',
'parent': 'FooOperator'},{
'children': [{
'children': [],
'class': 'C1Operator',
'parent': 'COperator'
},{
'children': [{
'children': [],
'class': 'C21Operator',
'parent': 'C2Operator'
}],
'class': 'C2Operator',
'parent': 'COperator'
}],
'class': 'COperator',
'parent': 'FooOperator'
}],
'class': 'FooOperator',
'parent': 'Operator'
}],
'class': 'Operator',
'parent': None
}
def display_tree(node, indent=0):
print('.' * indent, node['class'])
indent += 3
for child in node['children']:
display_tree(child, indent)
display_tree(tree)
您如何从
“C21Operator”
获取祖先列表,例如结果是[“Operator”、“FooOperator”、“COperator”、“C2Operator”、“C21Operator”]
?鉴于您的数据结构,我认为只有暴力解决方案是可能的:
In [6]: def path_to_child(tree, target, acc=None):
...: if acc is None:
...: acc = []
...: if tree['class'] == target:
...: return acc
...: for child in tree['children']:
...: found = path_to_child(child, target, acc + [tree['class']])
...: if found is not None:
...: return found
...:
In [7]: path_to_child(tree, 'C21Operator')
Out[7]: ['Operator', 'FooOperator', 'COperator', 'C2Operator']
In [8]:
如果您知道目标在哪里,您可能可以更智能地遍历树。您尝试了什么,它到底有什么问题?使用这种数据结构,我认为这是不可能的。好吧,如果你从
树开始走每一条可能的路径,然后返回指向的路径“C210Operator”
。但是,也许可以使用parent
属性实现您自己的节点
类,然后沿着父链?+1走到@juanpa。arrivillaga建议的自定义类实现更适合此问题的更好的数据结构。您的评论是对的,最好的方法是建立一个合适的数据结构,允许您线性地遍历父对象。总之,你解决了我想知道的问题;-)