<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 = defaultdict(TrieNode)
        self.word = None

    def addWord(self, word):
        cur = self
        for c in word:
            cur = cur.children[c]
        cur.word = word

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

        tree_root = {}
        def addNode(word):
            node = tree_root
            for c in word:
                if c not in node: node[c] = {}
                node = node[c]
            node["word"] = word

        m, n = len(board), len(board[0])
        DIR = [0, 1, 0, -1, 0]
        trieNode = TrieNode()
        ans = []
        for word in words:
            trieNode.addWord(word)
            addNode(word)

        def dfs(r, c, cur):
            if r < 0 or r == m or c < 0 or c == n or board[r][c] not in cur.children: return
            orgChar = board[r][c]
            cur = cur.children[orgChar]
            board[r][c] = '#'  # Mark as visited
            if cur.word != None:
                ans.append(cur.word)
                cur.word = None  # Avoid duplication!
            for i in range(4): dfs(r + DIR[i], c + DIR[i + 1], cur)
            board[r][c] = orgChar  # Restore to org state

        for r in range(m):
            for c in range(n):
                dfs(r, c, trieNode)
        return ans
19400