<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.endOfWord = False
    
    def addWord(self, word: str):
        curr = self
        for c in word:
            if c not in curr.children:
                curr.children[c] = TrieNode()
            curr = curr.children[c]
        curr.endOfWord = True


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        root = TrieNode()
        for w in words:
            root.addWord(w)

        ROWS, COLS = len(board), len(board[0])
        visited = set()
        res = set()
        
        def dfs(r: int, c: int, currWord: str, node: TrieNode):
            if (r < 0 or c < 0 or
                r == ROWS or c == COLS or
                board[r][c] not in node.children or
                (r, c) in visited):
                return
            
            char = board[r][c]            
            node = node.children[char]
            currWord += char
            if node.endOfWord:
                res.add(currWord)

            visited.add((r, c))
            dfs(r + 1, c, currWord, node)
            dfs(r - 1, c, currWord, node)
            dfs(r, c + 1, currWord, node)
            dfs(r, c - 1, currWord, node)
            visited.remove((r, c))
            return

        for r in range(ROWS):
            for c in range(COLS):
                dfs(r, c, "", root)
        
        return res

        
        
18600