<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>

struct TrieNode {
    string word = "";
    unordered_map<char, TrieNode*> letters;
};

class Solution {
public:
    vector<vector<int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    vector<vector<char>> board;

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        this->board = board;
        TrieNode* root = new TrieNode();
        vector<string> result;
        int rows = board.size();
        int cols = board[0].size();
        for (const string& word : words)
        {
            TrieNode* curr = root;
            for (const char letter : word)
            {
                if (curr->letters.find(letter) == curr->letters.end())
                {
                    TrieNode* newNode = new TrieNode();
                    curr->letters[letter] = newNode;
                }
                curr = curr->letters[letter];
            }
            curr->word = word;
        }

        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                if (root->letters.find(board[i][j]) != root->letters.end())
                {
                    dfs(i, j, root, result, "");
                }
            }
        }

        return result;
    }

    void dfs(int row, int col, TrieNode* curr, vector<string>& result, string word)
    {
        char temp = board[row][col];
        board[row][col] = '#';
        word += temp;
        TrieNode* child = curr->letters[temp];
        if (child->word != "")
        {
            result.push_back(child->word);
            child->word = "";
        }
        
        for (const auto& direction : directions)
        {
            int new_row = row + direction[0];
            int new_col = col + direction[1];

            if (new_row >= 0 && new_row < board.size() &&
                new_col >= 0 && new_col < board[0].size() &&
                child->letters.find(board[new_row][new_col]) != child->letters.end())
            {
                dfs(new_row, new_col, child, result, word + temp);
            }
        }
        board[row][col] = temp;
        if (child->letters.empty())
        {
            curr->letters.erase(temp);
        }
    }
};
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