<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.end = False
        self.seen = False

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

        self.root = Node()

        def populate():
            for word in words:
                trie = self.root
                for w in word:
                    if w not in trie.children:
                        trie.children[w] = Node()
                    trie = trie.children[w]
                trie.end = True
        
        populate()

        def dfs(r,c,trie):
            temp_arr.append(board[r][c])
        
            if trie.end and not trie.seen:
                res.append(''.join(temp_arr[:]))
                trie.seen = True
                
            temp = board[r][c]
            board[r][c] = '#'
            
            for nxtr, nxtc in [(0,1), (1,0),(-1,0),(0,-1)]:
                newr = r + nxtr
                newc = c + nxtc

                if (
                    0 <= newr < len(board)
                    and 0 <= newc < len(board[0])
                    and board[newr][newc] != '#'
                    and board[newr][newc] in trie.children
                ):
                    dfs(newr, newc, trie.children[board[newr][newc]])
            temp_arr.pop()
            board[r][c] = temp
        
        res = []
    
        trie = self.root

        for r in range(len(board)):
            for c in range(len(board[0])):
                if board[r][c] in trie.children:
                    temp_arr = []
                    dfs(r, c, trie.children[board[r][c]])
        return res

        
18900