您可以使用Python中的dataclasses模块创建Tree数据结构。

iter方法可用于使树可迭代，允许您通过改变yield语句的顺序来遍历树。

contains方法可用于检查树中是否存在特定值。

from dataclasses import dataclass

#               A
#              / \
#             B   C
#            / \   \
#           D   E   F
#          / \
#         G   H

@dataclass
class Node:
    data: str
    left: Node = None
    right: Node = None
    
    def __iter__(self):
        if self.left:
            yield from self.left
        
        yield self

        if self.right:
            yield from self.right

    def __contains__(self, other):
        for node in self:
            if node.data == other:
                return True
        return False
    

t = Node(
    'A', 
    Node(
        'B', 
        Node(
            'D', 
            Node('G'),
            Node('H'),
        ),
        Node('E'),
    ),  
    Node(
        'C', 
        right=Node('F'),
    ),
)
assert ('A' in t) is True
assert ('I' in t) is not True
for node in t:
    print(node.data, ' -> ', end='')
# G  -> D  -> H  -> B  -> E  -> A  -> C  -> F  -> 

Python不像Java那样具有相当广泛的“内置”数据结构。但是，因为Python是动态的，所以很容易创建通用树。例如，二叉树可能是:

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

你可以这样使用它:

root = Tree()
root.data = "root"
root.left = Tree()
root.left.data = "left"
root.right = Tree()
root.right.data = "right"

如果每个节点需要任意数量的子节点，则使用子节点列表:

class Tree:
    def __init__(self, data):
        self.children = []
        self.data = data

left = Tree("left")
middle = Tree("middle")
right = Tree("right")
root = Tree("root")
root.children = [left, middle, right]

class Tree(dict):
    """A tree implementation using python's autovivification feature."""
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

    #cast a (nested) dict to a (nested) Tree class
    def __init__(self, data={}):
        for k, data in data.items():
            if isinstance(data, dict):
                self[k] = type(self)(data)
            else:
                self[k] = data

作为一个字典，但提供尽可能多的嵌套字典。 试试下面的方法:

your_tree = Tree()

your_tree['a']['1']['x']  = '@'
your_tree['a']['1']['y']  = '#'
your_tree['a']['2']['x']  = '$'
your_tree['a']['3']       = '%'
your_tree['b']            = '*'

将传递一个嵌套的字典…就像树一样。

{'a': {'1': {'x': '@', 'y': '#'}, '2': {'x': '$'}, '3': '%'}, 'b': '*'}

．.． 如果你已经有字典了，它会把每一层都投射到一棵树上:

d = {'foo': {'amy': {'what': 'runs'} } }
tree = Tree(d)

print(d['foo']['amy']['what']) # returns 'runs'
d['foo']['amy']['when'] = 'now' # add new branch

这样，你就可以随心所欲地编辑/添加/删除每个词典级别。 遍历等所有dict方法仍然适用。

泛型树是一个具有零个或多个子节点的节点，每个子节点都是一个合适的(树)节点。它与二叉树不同，它们是不同的数据结构，尽管它们都有一些相同的术语。

Python中没有任何用于泛型树的内置数据结构，但很容易通过类实现。

class Tree(object):
    "Generic tree node."
    def __init__(self, name='root', children=None):
        self.name = name
        self.children = []
        if children is not None:
            for child in children:
                self.add_child(child)
    def __repr__(self):
        return self.name
    def add_child(self, node):
        assert isinstance(node, Tree)
        self.children.append(node)
#    *
#   /|\
#  1 2 +
#     / \
#    3   4
t = Tree('*', [Tree('1'),
               Tree('2'),
               Tree('+', [Tree('3'),
                          Tree('4')])])

我推荐任何树(我是作者)。

例子:

from anytree import Node, RenderTree

udo = Node("Udo")
marc = Node("Marc", parent=udo)
lian = Node("Lian", parent=marc)
dan = Node("Dan", parent=udo)
jet = Node("Jet", parent=dan)
jan = Node("Jan", parent=dan)
joe = Node("Joe", parent=dan)

print(udo)
Node('/Udo')
print(joe)
Node('/Udo/Dan/Joe')

for pre, fill, node in RenderTree(udo):
    print("%s%s" % (pre, node.name))
Udo
├── Marc
│   └── Lian
└── Dan
    ├── Jet
    ├── Jan
    └── Joe

print(dan.children)
(Node('/Udo/Dan/Jet'), Node('/Udo/Dan/Jan'), Node('/Udo/Dan/Joe'))

anytree也有一个强大的API:

简单的树创建 简单树修改 预序树迭代 后序树迭代 解析相对节点路径和绝对节点路径 从一个节点移动到另一个节点。 树渲染(参见上面的例子) 节点连接/分离连接

另一个基于Bruno回答的树的实现:

class Node:
    def __init__(self):
        self.name: str = ''
        self.children: List[Node] = []
        self.parent: Node = self

    def __getitem__(self, i: int) -> 'Node':
        return self.children[i]

    def add_child(self):
        child = Node()
        self.children.append(child)
        child.parent = self
        return child

    def __str__(self) -> str:
        def _get_character(x, left, right) -> str:
            if x < left:
                return '/'
            elif x >= right:
                return '\\'
            else:
                return '|'

        if len(self.children):
            children_lines: Sequence[List[str]] = list(map(lambda child: str(child).split('\n'), self.children))
            widths: Sequence[int] = list(map(lambda child_lines: len(child_lines[0]), children_lines))
            max_height: int = max(map(len, children_lines))
            total_width: int = sum(widths) + len(widths) - 1
            left: int = (total_width - len(self.name) + 1) // 2
            right: int = left + len(self.name)

            return '\n'.join((
                self.name.center(total_width),
                ' '.join(map(lambda width, position: _get_character(position - width // 2, left, right).center(width),
                             widths, accumulate(widths, add))),
                *map(
                    lambda row: ' '.join(map(
                        lambda child_lines: child_lines[row] if row < len(child_lines) else ' ' * len(child_lines[0]),
                        children_lines)),
                    range(max_height))))
        else:
            return self.name

还有一个如何使用它的例子:

tree = Node()
tree.name = 'Root node'
tree.add_child()
tree[0].name = 'Child node 0'
tree.add_child()
tree[1].name = 'Child node 1'
tree.add_child()
tree[2].name = 'Child node 2'
tree[1].add_child()
tree[1][0].name = 'Grandchild 1.0'
tree[2].add_child()
tree[2][0].name = 'Grandchild 2.0'
tree[2].add_child()
tree[2][1].name = 'Grandchild 2.1'
print(tree)

它应该输出:

                        Root node                        
     /             /                      \              
Child node 0  Child node 1           Child node 2        
                   |              /              \       
             Grandchild 1.0 Grandchild 2.0 Grandchild 2.1