<p>Given an <code>m x n</code> <code>board</code>&nbsp;of characters and a list of strings <code>words</code>, return <em>all words on the board</em>.</p>

<p>Each word must be constructed from letters of sequentially adjacent cells, where <strong>adjacent cells</strong> are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search1.jpg" style="width: 322px; height: 322px;" />
<pre>
<strong>Input:</strong> board = [[&quot;o&quot;,&quot;a&quot;,&quot;a&quot;,&quot;n&quot;],[&quot;e&quot;,&quot;t&quot;,&quot;a&quot;,&quot;e&quot;],[&quot;i&quot;,&quot;h&quot;,&quot;k&quot;,&quot;r&quot;],[&quot;i&quot;,&quot;f&quot;,&quot;l&quot;,&quot;v&quot;]], words = [&quot;oath&quot;,&quot;pea&quot;,&quot;eat&quot;,&quot;rain&quot;]
<strong>Output:</strong> [&quot;eat&quot;,&quot;oath&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> board = [[&quot;a&quot;,&quot;b&quot;],[&quot;c&quot;,&quot;d&quot;]], words = [&quot;abcb&quot;]
<strong>Output:</strong> []
</pre>

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

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 12</code></li>
	<li><code>board[i][j]</code> is a lowercase English letter.</li>
	<li><code>1 &lt;= words.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are unique.</li>
</ul>

class TrieNode:

    def __init__(self):
        self.children = {}
        self.end = False

class Trie:

    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        cur = self.root

        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()

            cur = cur.children[c]

        cur.end = True

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = Trie()

        for word in words:
            trie.insert(word)

        res = set()
        m, n = len(board), len(board[0])

        visited = [[False] * n for _ in range(m)]
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]

        def dfs(i, j, node, path):
            char = board[i][j]
            if char not in node.children:
                return
            
            visited[i][j] = True

            node = node.children[char]

            if node.end:
                res.add(path + char)
            
            for x, y in directions:
                ni, nj = i + x, j + y

                if 0 <= ni < m and 0 <= nj < n and not visited[ni][nj]:
                    dfs(ni, nj, node, path + char)

            visited[i][j] = False

        for i in range(m):
            for j in range(n):
                dfs(i, j, trie.root, "")

        return list(res)
18800