本文介绍 LeetCode 题集中，有关字典树的问题。
208. Implement Trie (Prefix Tree)（实现 Trie (前缀树)）
问题描述
思路与代码
本题是基本的字典树问题，使用 Python 语言中的字典类型实现比较方便。
代码如下：
class Trie:
def __init__(self):
self.trie = {
}
def insert(self, word: str) -> None:
cur = self.trie
for c in word:
if c not in cur.keys():
cur[c] = {
}
cur = cur[c]
cur['$'] = True
def search(self, word: str) -> bool:
cur = self.trie
for c in word:
if c not in cur.keys():
return False
cur = cur[c]
return '$' in cur.keys() # word must be the end of the tree, not just "start with"
def startsWith(self, prefix: str) -> bool:
cur = self.trie
for c in prefix:
if c not in cur.keys():
return False
cur = cur[c]
return True
# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.search(word)
# param_3 = obj.startsWith(prefix)
运行效果：
211. Design Add and Search Words Data Structure（添加与搜索单词 - 数据结构设计）
问题描述
思路与代码
本题与前一题相似，区别在于任意字符的匹配。
代码如下：
class WordDictionary:
def __init__(self):
self.dictionary = {
}
def addWord(self, word: str) -> None:
cur = self.dictionary
for c in word:
if c not in cur.keys():
cur[c] = {
}
cur = cur[c]
cur['$'] = {
} # can't be True like LeetCode 208, need to be iterable
def search(self, word: str) -> bool:
def rec(dic: Dict, i: int): # recursion function
if i == len(word):
return '$' in dic.keys()
if word[i] == '.':
for letter in dic.keys():
if rec(dic=dic[letter], i=i + 1):
return True
return False
elif word[i] not in dic.keys():
return False
else:
return rec(dic=dic[word[i]], i=i + 1)
return rec(dic=self.dictionary, i=0)
# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)
运行效果：
212. Word Search II（单词搜索 II）
问题描述
思路与代码
本题的思路，把单词列表写成字典树，然后在字母矩阵的不同路径的每个位置搜索是否存在当前单词。
代码如下：
class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
# create a trie
dictionary = {
}
for word in words:
dic = dictionary
for c in word:
if c not in dic.keys():
dic[c] = {
}
dic = dic[c]
dic['$'] = True
res = []
m, n = len(board), len(board[0])
def dfs(row: int, column: int, dic: Dict, path: str):
if '$' in dic.keys() and path not in res:
res.append(path)
mat_visit[row][column] = True
if row:
if not mat_visit[row - 1][column] and board[row - 1][column] in dic.keys():
dfs(row=row - 1, column=column, dic=dic[board[row - 1][column]], path=path + board[row - 1][column])
mat_visit[row - 1][column] = False
if row < m - 1:
if not mat_visit[row + 1][column] and board[row + 1][column] in dic.keys():
dfs(row=row + 1, column=column, dic=dic[board[row + 1][column]], path=path + board[row + 1][column])
mat_visit[row + 1][column] = False
if column:
if not mat_visit[row][column - 1] and board[row][column - 1] in dic.keys():
dfs(row=row, column=column - 1, dic=dic[board[row][column - 1]], path=path + board[row][column - 1])
mat_visit[row][column - 1] = False
if column < n - 1:
if not mat_visit[row][column + 1] and board[row][column + 1] in dic.keys():
dfs(row=row, column=column + 1, dic=dic[board[row][column + 1]], path=path + board[row][column + 1])
mat_visit[row][column + 1] = False
for i in range(m):
for j in range(n):
mat_visit = [[False for _ in range(n)] for _ in range(m)]
if board[i][j] in dictionary.keys():
dfs(row=i, column=j, dic=dictionary[board[i][j]], path=board[i][j])
return res
运行效果：
