我将根树实现为字典{child:parent}。比如根节点为0，树可能是这样的:

tree={1:0, 2:0, 3:1, 4:2, 5:3}

这种结构使得沿着一条路径从任意节点向上到根结点非常容易，这与我正在处理的问题有关。

我已经在我的网站https://web.archive.org/web/20120723175438/www.quesucede.com/page/show/id/python_3_tree_implementation上发布了一个Python 3树的实现

代码如下:

import uuid

def sanitize_id(id):
    return id.strip().replace(" ", "")

(_ADD, _DELETE, _INSERT) = range(3)
(_ROOT, _DEPTH, _WIDTH) = range(3)

class Node:

    def __init__(self, name, identifier=None, expanded=True):
        self.__identifier = (str(uuid.uuid1()) if identifier is None else
                sanitize_id(str(identifier)))
        self.name = name
        self.expanded = expanded
        self.__bpointer = None
        self.__fpointer = []

    @property
    def identifier(self):
        return self.__identifier

    @property
    def bpointer(self):
        return self.__bpointer

    @bpointer.setter
    def bpointer(self, value):
        if value is not None:
            self.__bpointer = sanitize_id(value)

    @property
    def fpointer(self):
        return self.__fpointer

    def update_fpointer(self, identifier, mode=_ADD):
        if mode is _ADD:
            self.__fpointer.append(sanitize_id(identifier))
        elif mode is _DELETE:
            self.__fpointer.remove(sanitize_id(identifier))
        elif mode is _INSERT:
            self.__fpointer = [sanitize_id(identifier)]

class Tree:

    def __init__(self):
        self.nodes = []

    def get_index(self, position):
        for index, node in enumerate(self.nodes):
            if node.identifier == position:
                break
        return index

    def create_node(self, name, identifier=None, parent=None):

        node = Node(name, identifier)
        self.nodes.append(node)
        self.__update_fpointer(parent, node.identifier, _ADD)
        node.bpointer = parent
        return node

    def show(self, position, level=_ROOT):
        queue = self[position].fpointer
        if level == _ROOT:
            print("{0} [{1}]".format(self[position].name,
                                     self[position].identifier))
        else:
            print("\t"*level, "{0} [{1}]".format(self[position].name,
                                                 self[position].identifier))
        if self[position].expanded:
            level += 1
            for element in queue:
                self.show(element, level)  # recursive call

    def expand_tree(self, position, mode=_DEPTH):
        # Python generator. Loosly based on an algorithm from 'Essential LISP' by
        # John R. Anderson, Albert T. Corbett, and Brian J. Reiser, page 239-241
        yield position
        queue = self[position].fpointer
        while queue:
            yield queue[0]
            expansion = self[queue[0]].fpointer
            if mode is _DEPTH:
                queue = expansion + queue[1:]  # depth-first
            elif mode is _WIDTH:
                queue = queue[1:] + expansion  # width-first

    def is_branch(self, position):
        return self[position].fpointer

    def __update_fpointer(self, position, identifier, mode):
        if position is None:
            return
        else:
            self[position].update_fpointer(identifier, mode)

    def __update_bpointer(self, position, identifier):
        self[position].bpointer = identifier

    def __getitem__(self, key):
        return self.nodes[self.get_index(key)]

    def __setitem__(self, key, item):
        self.nodes[self.get_index(key)] = item

    def __len__(self):
        return len(self.nodes)

    def __contains__(self, identifier):
        return [node.identifier for node in self.nodes
                if node.identifier is identifier]

if __name__ == "__main__":

    tree = Tree()
    tree.create_node("Harry", "harry")  # root node
    tree.create_node("Jane", "jane", parent = "harry")
    tree.create_node("Bill", "bill", parent = "harry")
    tree.create_node("Joe", "joe", parent = "jane")
    tree.create_node("Diane", "diane", parent = "jane")
    tree.create_node("George", "george", parent = "diane")
    tree.create_node("Mary", "mary", parent = "diane")
    tree.create_node("Jill", "jill", parent = "george")
    tree.create_node("Carol", "carol", parent = "jill")
    tree.create_node("Grace", "grace", parent = "bill")
    tree.create_node("Mark", "mark", parent = "jane")

    print("="*80)
    tree.show("harry")
    print("="*80)
    for node in tree.expand_tree("harry", mode=_WIDTH):
        print(node)
    print("="*80)

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  -> 

我使用嵌套字典实现了树。这很容易做到，而且对我来说，它在相当大的数据集上很有效。我在下面发布了一个示例，你可以在谷歌代码中看到更多

  def addBallotToTree(self, tree, ballotIndex, ballot=""):
    """Add one ballot to the tree.

    The root of the tree is a dictionary that has as keys the indicies of all 
    continuing and winning candidates.  For each candidate, the value is also
    a dictionary, and the keys of that dictionary include "n" and "bi".
    tree[c]["n"] is the number of ballots that rank candidate c first.
    tree[c]["bi"] is a list of ballot indices where the ballots rank c first.

    If candidate c is a winning candidate, then that portion of the tree is
    expanded to indicate the breakdown of the subsequently ranked candidates.
    In this situation, additional keys are added to the tree[c] dictionary
    corresponding to subsequently ranked candidates.
    tree[c]["n"] is the number of ballots that rank candidate c first.
    tree[c]["bi"] is a list of ballot indices where the ballots rank c first.
    tree[c][d]["n"] is the number of ballots that rank c first and d second.
    tree[c][d]["bi"] is a list of the corresponding ballot indices.

    Where the second ranked candidates is also a winner, then the tree is 
    expanded to the next level.  

    Losing candidates are ignored and treated as if they do not appear on the 
    ballots.  For example, tree[c][d]["n"] is the total number of ballots
    where candidate c is the first non-losing candidate, c is a winner, and
    d is the next non-losing candidate.  This will include the following
    ballots, where x represents a losing candidate:
    [c d]
    [x c d]
    [c x d]
    [x c x x d]

    During the count, the tree is dynamically updated as candidates change
    their status.  The parameter "tree" to this method may be the root of the
    tree or may be a sub-tree.
    """

    if ballot == "":
      # Add the complete ballot to the tree
      weight, ballot = self.b.getWeightedBallot(ballotIndex)
    else:
      # When ballot is not "", we are adding a truncated ballot to the tree,
      # because a higher-ranked candidate is a winner.
      weight = self.b.getWeight(ballotIndex)

    # Get the top choice among candidates still in the running
    # Note that we can't use Ballots.getTopChoiceFromWeightedBallot since
    # we are looking for the top choice over a truncated ballot.
    for c in ballot:
      if c in self.continuing | self.winners:
        break # c is the top choice so stop
    else:
      c = None # no candidates left on this ballot

    if c is None:
      # This will happen if the ballot contains only winning and losing
      # candidates.  The ballot index will not need to be transferred
      # again so it can be thrown away.
      return

    # Create space if necessary.
    if not tree.has_key(c):
      tree[c] = {}
      tree[c]["n"] = 0
      tree[c]["bi"] = []

    tree[c]["n"] += weight

    if c in self.winners:
      # Because candidate is a winner, a portion of the ballot goes to
      # the next candidate.  Pass on a truncated ballot so that the same
      # candidate doesn't get counted twice.
      i = ballot.index(c)
      ballot2 = ballot[i+1:]
      self.addBallotToTree(tree[c], ballotIndex, ballot2)
    else:
      # Candidate is in continuing so we stop here.
      tree[c]["bi"].append(ballotIndex)

我将根树实现为字典{child:parent}。比如根节点为0，树可能是这样的:

tree={1:0, 2:0, 3:1, 4:2, 5:3}

这种结构使得沿着一条路径从任意节点向上到根结点非常容易，这与我正在处理的问题有关。

class Node:
    """
    Class Node
    """
    def __init__(self, value):
        self.left = None
        self.data = value
        self.right = None

class Tree:
    """
    Class tree will provide a tree as well as utility functions.
    """

    def createNode(self, data):
        """
        Utility function to create a node.
        """
        return Node(data)

    def insert(self, node , data):
        """
        Insert function will insert a node into tree.
        Duplicate keys are not allowed.
        """
        #if tree is empty , return a root node
        if node is None:
            return self.createNode(data)
        # if data is smaller than parent , insert it into left side
        if data < node.data:
            node.left = self.insert(node.left, data)
        elif data > node.data:
            node.right = self.insert(node.right, data)

        return node


    def search(self, node, data):
        """
        Search function will search a node into tree.
        """
        # if root is None or root is the search data.
        if node is None or node.data == data:
            return node

        if node.data < data:
            return self.search(node.right, data)
        else:
            return self.search(node.left, data)



    def deleteNode(self,node,data):
        """
        Delete function will delete a node into tree.
        Not complete , may need some more scenarion that we can handle
        Now it is handling only leaf.
        """

        # Check if tree is empty.
        if node is None:
            return None

        # searching key into BST.
        if data < node.data:
            node.left = self.deleteNode(node.left, data)
        elif data > node.data:
            node.right = self.deleteNode(node.right, data)
        else: # reach to the node that need to delete from BST.
            if node.left is None and node.right is None:
                del node
            if node.left == None:
                temp = node.right
                del node
                return  temp
            elif node.right == None:
                temp = node.left
                del node
                return temp

        return node

    def traverseInorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            self.traverseInorder(root.left)
            print(root.data)
            self.traverseInorder(root.right)

    def traversePreorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            print(root.data)
            self.traversePreorder(root.left)
            self.traversePreorder(root.right)

    def traversePostorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            self.traversePostorder(root.left)
            self.traversePostorder(root.right)
            print(root.data)


def main():
    root = None
    tree = Tree()
    root = tree.insert(root, 10)
    print(root)
    tree.insert(root, 20)
    tree.insert(root, 30)
    tree.insert(root, 40)
    tree.insert(root, 70)
    tree.insert(root, 60)
    tree.insert(root, 80)

    print("Traverse Inorder")
    tree.traverseInorder(root)

    print("Traverse Preorder")
    tree.traversePreorder(root)

    print("Traverse Postorder")
    tree.traversePostorder(root)


if __name__ == "__main__":
    main()