在python中遍历树层次结构？

Question

我是新来的，对python很陌生！

我们有家庭作业，我已经可以做剩下的了，但还有一个问题： 如果我有这样的树层次结构：

root = [
    parent1 = [
        child1,
        child2 = [
            sub_child
        ]
        child3
    ],
    parent2 = [
        child1,
        child2
    ]
]

它们都是一个名为TreeHierarchyClass的类的实例，它们都有一个name属性，我如何找到输入了name的那个？

我试着用循环，但没办法知道我需要多少？取名字很容易：

name = input("Enter name: ")
if name == TreeHierarchyObject.name:
    print("Found it!")

但我该如何循环遍历对象呢？

Answer 1

您可以使用recursion，也可以使用iteration。无论哪种方式都无关紧要。但是你需要一个搜索树的策略。

以下是一些策略，可用于查看图表：

• http://en.wikipedia.org/wiki/Breadth-first_search
• http://en.wikipedia.org/wiki/Depth-first_search

其主要思想是不经过同一个节点/叶两次，这对于树来说是微不足道的，但是对于图来说需要coloring：

• http://en.wikipedia.org/wiki/Graph_coloring

有一些设计模式可以使用，例如visitor模式，您可以向.visit()添加方法TreeHierarchyClass来访问其子节点，并添加另一个方法来按名称查找节点。

• http://en.wikipedia.org/wiki/Visitor_pattern

示例：

# imagine we got this class
class TreeHierarchyClass(object):
    def __init__(self, value):
        self.children = []
        self.value = value
        if self.value == 13:
            self.name = 'the lucky one.'
    def add(self, value):
        self.children.append(type(self)(value))

您可以通过以下方式访问所有节点：

def visit(tree):
    visited = set()
    nonvisited = set()
    nonvisited.update(tree.children)
    while nonvisited:
        item = nonvisited.pop()
        # already seen
        if item in visited:
            continue
        # mark item
        visited.add(item)
        yield item
        # add children
        nonvisited.update(item.children)

让我们构建一个示例树结构：

root = TreeHierarchyClass(0)

for i in range(10):
    root.add(i)

for i in range(10):
    root.children[1].add(i + 10)

现在让我们找到一些项目：

def find(name):
    for item in visit(root):
        print 'checking item with value %d' % item.value,
        if getattr(item, 'name', None) == name:
            print '- found it.'
            break
        else:
            print '- nope, keep searching.'
    else:
        print 'Sorry, not found.'

find('the lucky one.')
find('the lost one.')

此示例将打印：

>>> find('the lucky one.')
checking item with value 7 - nope, keep searching.
checking item with value 0 - nope, keep searching.
checking item with value 1 - nope, keep searching.
checking item with value 12 - nope, keep searching.
checking item with value 2 - nope, keep searching.
checking item with value 9 - nope, keep searching.
checking item with value 19 - nope, keep searching.
checking item with value 3 - nope, keep searching.
checking item with value 11 - nope, keep searching.
checking item with value 4 - nope, keep searching.
checking item with value 14 - nope, keep searching.
checking item with value 5 - nope, keep searching.
checking item with value 6 - nope, keep searching.
checking item with value 15 - nope, keep searching.
checking item with value 8 - nope, keep searching.
checking item with value 16 - nope, keep searching.
checking item with value 13 - found it.
>>> find('the lost one.')
checking item with value 7 - nope, keep searching.
checking item with value 0 - nope, keep searching.
checking item with value 1 - nope, keep searching.
checking item with value 12 - nope, keep searching.
checking item with value 2 - nope, keep searching.
checking item with value 9 - nope, keep searching.
checking item with value 19 - nope, keep searching.
checking item with value 3 - nope, keep searching.
checking item with value 11 - nope, keep searching.
checking item with value 4 - nope, keep searching.
checking item with value 14 - nope, keep searching.
checking item with value 5 - nope, keep searching.
checking item with value 6 - nope, keep searching.
checking item with value 15 - nope, keep searching.
checking item with value 8 - nope, keep searching.
checking item with value 16 - nope, keep searching.
checking item with value 13 - nope, keep searching.
checking item with value 17 - nope, keep searching.
checking item with value 10 - nope, keep searching.
checking item with value 18 - nope, keep searching.
Sorry, not found.

Answer 2

你应该在这里使用简单的递归。 该方法稍微取决于子对象如何附加到父对象。

如果它们在列表self.children中（我建议您这样做），则此方法有效。 只需在类中定义以下方法：

def findObjectByName(self, name):
    if self.name == name:
        return self
    else:
        for child in self.children:
            match = child.findObjectByName(name)
            if match:
                return match

编辑： 要使此操作适用于任何属性，而不仅仅是名称，请改用getattr()：

def findObject(self, attr, value):
    if getattr(self, attr) == value:
        return self
    else:
        for child in self.children:
            match = child.findObject(attr, value)
            if match:
                return match

只需调用root.findObjectByName("Sub Child!")或使用第二种方法：root.findObject("name", "Sub Child!")