<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{
public:
    vector<TrieNode*> children;
    string word;
    bool is_word;
    TrieNode(): word(""), is_word(false) {
        for(int i = 0; i < 26; i++) {
            children.push_back(NULL);
        }
    }
};
class Solution {
private:
    TrieNode* root = new TrieNode();
public:
    void buildTree(vector<string>& words) {
        for(auto word : words) {
            TrieNode* p = root;
            for(auto c : word) {
                int index = c - 'a';
                if(p->children[index] == NULL) {
                    p->children[index] = new TrieNode();
                }
                p = p->children[index];
            }
            p->is_word = true;
            p->word = word;
        }
    }
    vector<vector<int>> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; 
    bool dfs(vector<string>& res, TrieNode* root, vector<vector<char>>& board, int i, int j, int n, int m) {
        if(root == NULL) return true;
        char tmp = board[i][j];
        board[i][j] = '0';
        
        if(root->is_word) {
            res.push_back(root->word);
            // incase duplicate;
            root->is_word = false;
            root->word = "";
        }
        for(auto d : dir) {
            int x = i + d[0], y = j + d[1];
            if(x < 0 || x >= n || y < 0 || y >= m || board[x][y] == '0') continue;
            int index = board[x][y] - 'a';
            if(dfs(res, root->children[index], board, x, y, n, m)) {
                root->children[index] = NULL;
            }; 
        }
        board[i][j] = tmp;
        for(int i = 0; i < 26; i++) {
            if(root->children[i] != NULL) return false;
        }
        return true;

    }
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        vector<string> res;

        // buildTree;
        buildTree(words);

        //traverse Tree and board;
        int n = board.size(), m = board[0].size();
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                dfs(res, root->children[board[i][j] - 'a'], board, i, j, n, m);
            }
        }

        return res;
    }
};
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