<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>

<ul>
	<li>Every adjacent pair of words differs by a single letter.</li>
	<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
	<li><code>s<sub>k</sub> == endWord</code></li>
</ul>

<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>all the <strong>shortest transformation sequences</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]
<strong>Output:</strong> [[&quot;hit&quot;,&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;cog&quot;],[&quot;hit&quot;,&quot;hot&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]]
<strong>Explanation:</strong>&nbsp;There are 2 shortest transformation sequences:
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; &quot;cog&quot;
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;lot&quot; -&gt; &quot;log&quot; -&gt; &quot;cog&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]
<strong>Output:</strong> []
<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= beginWord.length &lt;= 5</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 500</code></li>
	<li><code>wordList[i].length == beginWord.length</code></li>
	<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
	<li><code>beginWord != endWord</code></li>
	<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
	<li>The <strong>sum</strong> of all shortest transformation sequences does not exceed <code>10<sup>5</sup></code>.</li>
</ul>

class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        # 这个实现的关键在于 BFS 确保找到的是最短路径，然后 DFS 反向构建这些路径
        ans = []
        if endWord not in wordList:
            return ans
        size = len(beginWord)
        cur_word_set = set()
        cur_word_set.add(beginWord)
        ws = set(wordList)
        # 用于控制bfs结束
        flag = False
        if beginWord in ws:
             ws.remove(beginWord)
        tmp = list('abcdefghijklmnopqrstuvwxyz')
        # key为当前单词，value为能到达该单词（bfs搜索）的单词列表，即该单词的parent
        word_dict = {}
        # bfs
        while cur_word_set:
            # 该次bfs匹配到的单词
            match_set = set()
            for cur in cur_word_set:
                for i in range(size):
                    for j in tmp:
                        if cur[i] == j:
                            continue
                        word = cur[:i] + j + cur[i + 1:]
                        # word在单词列表中
                        if word in ws:
                            # 匹配加入该单词
                            match_set.add(word)
                            # 为该单词维护parent
                            if word not in word_dict:
                                word_dict[word] = []
                            word_dict[word].append(cur)
                            # 到达终点 bfs结束
                            if word == endWord:
                                flag = True
            # 该次bfs结束，剔除掉匹配到的单词
            for el in match_set:
                ws.remove(el)
            # 下次bfs的parent
            cur_word_set = match_set
            if flag: break
        # 完成bfs开始dfs,通过endword反向找parent,直到找到beginword

        # 没有匹到endword,直接结束
        if not flag: return ans
        result = [endWord]
        def dfs(cur_word):
            if cur_word == beginWord:
                ans.append(result[::-1])
                return
            for el in word_dict[cur_word]:
                result.append(el)
                dfs(el)
                result.pop()

        dfs(endWord)
        return ans


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