<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, char):
        self.char = char
        self.children = dict() 
        self.word = ''

class Trie:

    def __init__(self):
        self.root = TrieNode('')
    
    def add_word(self, word):
        curr = self.root
        for c in word:
            if c not in curr.children:
                curr.children[c] = TrieNode(c)
            curr = curr.children[c]
        curr.word = word

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        m = len(board)
        n = len(board[0])

        trie = Trie()
        for word in words:
            if len(word) > m * n:
                continue 
            trie.add_word(word)
        
        res = []

        offsets = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        seen = set() 

        def dfs(row, col, tnode):
            # row, col already in seen
            # board[row][col] == tnode.char 
            
            if tnode.word:
                res.append(tnode.word)
                tnode.word = ""
            
            for dr, dc in offsets:
                nr = row + dr
                nc = col + dc
                if 0 <= nr < len(board) and 0 <= nc < len(board[0]) and (nr, nc) not in seen and board[nr][nc] in tnode.children:
                    seen.add((nr, nc))
                    dfs(nr, nc, tnode.children[board[nr][nc]])
                    seen.remove((nr, nc))
        
        for r in range(len(board)):
            for c in range(len(board[0])):
                if board[r][c] in trie.root.children:
                    seen.add((r, c))
                    dfs(r, c, trie.root.children[board[r][c]])
                    seen.remove((r, c))
        
        return res






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