本题要求的是最短转换序列的长度，看到最短首先想到的就是广度优先搜索。想到广度优先搜索自然而然的就能想到图，但是本题并没有直截了当的给出图的模型，因此需要按照题意把它抽象成图的模型。

python">class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:        
        edge = dict() # 记录无向图中每个单词的连接单词
        visited = dict() # 记录单词是否被访问过，避免死循环
        res = 1 # beginWord本身为一个长度

        # 快速查找endWord是否在wordList中
        for word in wordList:
            edge[word] = []
        if endWord not in edge:
            return 0
        # 将beginWord纳入wordList中，以在后续建图过程中将其划入图中
        if beginWord not in edge:
            edge[beginWord] = []
            wordList = [beginWord] + wordList

        # 建图时间复杂度O(N*C*26)，比两两单词判断(O(N*N*C))快
        for word in wordList:
            list_word = list(word)
            n = len(list_word)
            for i in range(n):
                tmp = list_word[i]
                for k in range(26):
                    list_word[i] = chr(ord('a') + k)
                    newWord = ''.join(list_word)
                    if newWord in edge and newWord != word:
                        edge[word].append(newWord)
                list_word[i] = tmp
        
        queue = [beginWord] # 以beginWord为起点，开始BFS过程

        while queue:
            current = len(queue)
            for _ in range(current):
                node = queue[0]
                queue.pop(0)
                visited[node] = True
                for vertex in edge[node]:
                    if vertex == endWord:
                        return res + 1
                    if vertex not in visited:
                        queue.append(vertex)
            res += 1
                       
        return 0