我已经阅读了 python 中 trie 的以下实现： https://stackoverflow.com/a/11016430/2225221
并尝试为其设置删除功能。 基本上，我一开始就有问题：如果你想从树中删除一个词，它可以有子“词”，也可以是另一个词的“子词”。
如果你用“del dict[key]”删除，你也删除了上面提到的这两种词。 任何人都可以帮我解决这个问题，如何正确删除所选单词（让我们假设它在 trie 中）
【问题讨论】：
标签： python python-2.7 trie
基本上，要从特里删除一个单词（因为它在您链接到的答案中实现），您只需要删除它的 _end 标记，例如：
def remove_word(trie, word):
current_dict = trie
for letter in word:
current_dict = current_dict.get(letter, None)
if current_dict is None:
# the trie doesn't contain this word.
break
else:
del current_dict[_end]
但是请注意，这并不能确保 trie 具有最小的大小。删除单词后，可能会在左侧的 trie 中存在不再被任何单词使用的分支。这不会影响数据结构的正确性，它只是意味着 trie 可能会消耗比绝对必要更多的内存。您可以通过从叶节点向后迭代并删除分支来改进这一点，直到找到一个具有多个子节点的分支。
编辑： 这里有一个想法，你可以如何实现一个删除函数，该函数还剔除任何不必要的分支。可能有一种更有效的方法，但这可能会让你开始：
def remove_word2(trie, word):
current_dict = trie
path = [current_dict]
for letter in word:
current_dict = current_dict.get(letter, None)
path.append(current_dict)
if current_dict is None:
# the trie doesn't contain this word.
break
else:
if not path[-1].get(_end, None):
# the trie doesn't contain this word (but a prefix of it).
return
deleted_branches = []
for current_dict, letter in zip(reversed(path[:-1]), reversed(word)):
if len(current_dict[letter]) <= 1:
deleted_branches.append((current_dict, letter))
else:
break
if len(deleted_branches) > 0:
del deleted_branches[-1][0][deleted_branches[-1][1]]
del path[-1][_end]
本质上，它首先找到将要删除的单词的“路径”，然后向后迭代以找到可以删除的节点。然后它会删除可以删除的路径的根（这也隐式删除了_end 节点）。
【讨论】：
remove_word2。如果len(current_dict) > 1（替换任何以前保存的值），您可以只保存当前字典和字母，而不是总是将当前字典添加到您的path 列表中。然后在循环结束后，从保存的字典中删除保存的字母。
我觉得最好是递归的，代码如下：
def remove(self, word):
self.delete(self.tries, word, 0)
def delete(self, dicts, word, i):
if i == len(word):
if 'end' in dicts:
del dicts['end']
if len(dicts) == 0:
return True
else:
return False
else:
return False
else:
if word[i] in dicts and self.delete(dicts[word[i]], word, i + 1):
if len(dicts[word[i]]) == 0:
del dicts[word[i]]
return True
else:
return False
else:
return False
【讨论】：
def remove_a_word_util(self, word, idx, node):
if len(word) == idx:
node.is_end_of_word = False
return bool(node.children)
ch = word[idx]
if ch not in node.children:
return True
flag = self.remove_a_word_util(word, idx+1, node.children[ch])
if flag:
return True
node.children.pop(ch)
return bool(node.children) or node.is_end_of_word
【讨论】：
处理此类结构的一种方法是通过recursion。在这种情况下，递归的好处在于它会拉到 trie 的底部，然后将返回的值通过分支传递回来。
下面的函数就是这样做的。它转到叶子并删除_end 值，以防输入单词是另一个单词的前缀。然后它传递一个布尔值 (boo)，表示 current_dict 仍位于外围分支中。一旦我们达到当前 dict 有多个子节点的点，我们删除相应的分支并将 boo 设置为 False，这样每个剩余的递归都不会执行任何操作。
def trie_trim(term, trie=SYNONYMS, prev=0):
# checks that we haven't hit the end of the word
if term:
first, rest = term[0], term[1:]
current_length = len(trie)
next_length, boo = trie_trim(rest, trie=trie[first], prev=current_length)
# this statement avoids trimming excessively if the input is a prefix because
# if the word is a prefix, the first returned value will be greater than 1
if boo and next_length > 1:
boo = False
# this statement checks for the first occurrence of the current dict having more than one child
# or it checks that we've hit the bottom without trimming anything
elif boo and (current_length > 1 or not prev):
del trie[first]
boo = False
return current_length, boo
# when we do hit the end of the word, delete _end
else:
del trie[_end]
return len(trie) + 1, True
【讨论】：
有点长，但我希望这有助于回答您的问题：
class Trie:
WORD_END = "$"
def __init__(self):
self.trie = {}
def insert(self, word):
cur = self.trie
for char in word:
if char not in cur:
cur[char] = {}
cur = cur[char]
cur[Trie.WORD_END] = word
def delete(self, word):
def _delete(word, cur_trie, i=0):
if i == len(word):
if Trie.WORD_END not in cur_trie:
raise ValueError("'%s' is not registered in the trie..." %word)
cur_trie.pop(Trie.WORD_END)
if len(cur_trie) > 0:
return False
return True
if word[i] not in cur_trie:
raise ValueError("'%s' is not registered in the trie..." %word)
cont = _delete(word, cur_trie[word[i]], i+1)
if cont:
cur_trie.pop(word[i])
if Trie.WORD_END in cur_trie:
return False
return True
return False
_delete(word, self.trie)
t = Trie()
t.insert("bar")
t.insert("baraka")
t.insert("barakalar")
t.delete("barak") # raises error as 'barak' is not a valid WORD_END although it is a valid path.
t.delete("bareka") # raises error as 'e' does not exist in the path.
t.delete("baraka") # deletes the WORD_END of 'baraka' without deleting any letter as there is 'barakalar' afterwards.
t.delete("barakalar") # deletes until the previous word (until the first Trie.WORD_END; "$" - by going backwards with recursion) in the same path (until 'baraka').
【讨论】：
如果您需要整个 DS：
class TrieNode:
def __init__(self):
self.children = {}
self.wordCounter = 0
self.prefixCounter = 0
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, word: str) -> None:
node = self.root
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node.prefixCounter += 1
node = node.children[char]
node.wordCounter += 1
def countWordsEqualTo(self, word: str) -> int:
node = self.root
if node.children:
for char in word:
node = node.children[char]
else:
return 0
return node.wordCounter
def countWordsStartingWith(self, prefix: str) -> int:
node = self.root
if node.children:
for char in prefix:
node = node.children[char]
else:
return 0
return node.prefixCounter
def erase(self, word: str) -> None:
node = self.root
for char in word:
if node.children:
node.prefixCounter -= 1
node = node.children[char]
else:
return None
node.wordCounter -= 1
if node.wordCounter == 0:
self.dfsRemove(self.root, word, 0)
def dfsRemove(self, node: TrieNode, word: str, idx: int) -> None:
if len(word) == idx:
node.wordCounter = 0
return
char = word[idx]
if char not in node.children:
return
self.dfsRemove(node.children[char], word, idx+1)
node.children.pop(char)
trie = Trie();
trie.insert("apple"); #// Inserts "apple".
trie.insert("apple"); #// Inserts another "apple".
print(trie.countWordsEqualTo("apple")) #// There are two instances of "apple" so return 2.
print(trie.countWordsStartingWith("app")) #// "app" is a prefix of "apple" so return 2.
trie.erase("apple") #// Erases one "apple".
print(trie.countWordsEqualTo("apple")) #// Now there is only one instance of "apple" so return 1.
print(trie.countWordsStartingWith("app")) #// return 1
trie.erase("apple"); #// Erases "apple". Now the trie is empty.
print(trie.countWordsEqualTo("apple")) #// return 0
print(trie.countWordsStartingWith("app")) #// return 0
【讨论】：