<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 Node:
    def __init__(self):
        self.children = {}
        self.word = False


class Trie:
    def __init__(self):
        self.root = Node()

    def insert(self, word: str) -> None:
        root = self.root

        for ch in word:
            if ch not in root.children:
                root.children[ch] = Node()
            root = root.children[ch]

        root.word = True   


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        self.trie = Trie()
        self.rows = len(board)
        self.cols = len(board[0])
        self.ans = {}

        for word in words:
            self.trie.insert(word)

        def dfs(board, i, j, path, root):
            ch = board[i][j]

            # Check if current character is in trie
            root = root.children.get(ch)
            if root is None:
                return

            # Update path and mark visited
            path = path + [ch]
            board[i][j] = None

            # Add to solution if node is marked as a word in trie
            if root.word:
                word = ''.join(path)
                self.ans[word] = True

            # Right
            if j + 1 < self.cols and board[i][j+1]:
                dfs(board, i, j+1, path, root)

            # Left
            if j - 1 >= 0 and board[i][j-1]:
                dfs(board, i, j-1, path, root)

            # Down
            if i + 1 < self.rows and board[i+1][j]:
                dfs(board, i+1, j, path, root)

            # Up
            if i - 1 >= 0 and board[i-1][j]:
                dfs(board, i-1, j, path, root)
            
            board[i][j] = ch
        
        for i in range(self.rows):
            for j in range(self.cols):
                dfs(board, i, j, [], self.trie.root)

        return self.ans.keys()
        
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