<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.isWord = False
class Trie:
    def __init__(self):
        self.root = TrieNode()
    def insert(self, word):
        cur = self.root
        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()
            cur = cur.children[c]
        cur.isWord = True 
    def remove(self, word):
        cur = self.root
        nodes = [cur]
        for c in word:
            cur = cur.children[c]
            nodes.append(cur)
        nodes[-1].isWord = False
        
        for i in range(len(word)-1, -1, -1):
            parent = nodes[i]
            char = word[i]
            child = parent.children[char]
            if not child.children and not child.isWord:
                del parent.children[char]
            else:
                break
    
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        ROW, COL = len(board), len(board[0])
        trie = Trie()
        firstCharSet = set()
        visited = set()
        for w in words:
            firstCharSet.add(w[0])
            trie.insert(w)
        
        res = []
        def dfs(r, c, cur, curWord):
            if r not in range(ROW) or c not in range(COL) or (r,c) in visited or board[r][c] not in cur.children:
                return 
            visited.add((r,c))
            curWord.append(board[r][c])
            cur = cur.children[board[r][c]]
            if cur.isWord:
                res.append("".join(curWord))
                trie.remove("".join(curWord))
            dfs(r-1, c, cur, curWord)
            dfs(r+1, c, cur, curWord)
            dfs(r, c-1, cur, curWord)
            dfs(r, c+1, cur, curWord)
            visited.remove((r,c))
            curWord.pop()


        for r in range(ROW):
            for c in range(COL):
                if board[r][c] in firstCharSet:
                    cur = trie.root
                    curWord = []
                    dfs(r, c, cur, curWord)

        return res
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