如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
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()
其他回答
如果有人需要一个更简单的方法,树只是一个递归嵌套的列表(因为set是不可哈希的):
[root, [child_1, [[child_11, []], [child_12, []]], [child_2, []]]]
每个分支都是一对:[object, [children]] 每个叶子是一对:[object, []]
但是如果你需要一个带有方法的类,你可以使用任何树。
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。
你可以试试:
from collections import defaultdict
def tree(): return defaultdict(tree)
users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'
建议在这里:https://gist.github.com/2012250
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)
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