【发布时间】:2017-02-27 22:41:42
【问题描述】:
给定一棵树,其中每个节点可以有 N 个子节点,但只有 1 个父节点。如何获得一个节点的祖先?例如,假设我得到了这棵树:
# Operator
# ... FooOperator
# ...... BOperator
# ......... B1Operator
# ............ B11Operator
# ...... AOperator
# ......... A2Operator
# ......... A1Operator
# ......... A3Operator
# ...... 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,例如结果为["Operator", "FooOperator", "COperator", "C2Operator", "C21Operator"]?
【问题讨论】:
-
你尝试过什么,它到底有什么问题?
-
使用这种数据结构,我认为这不太可能。好吧,也许如果你从
tree开始走所有可能的路径,然后返回引导你到"C210Operator"的路径。但也许用parent属性实现你自己的Node类,然后走父链? -
+1 对@juanpa.arrivillaga 的自定义类的建议,该类实现了更适合此问题的更好的数据结构。
标签: python python-3.x tree ancestor