<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{
    unordered_map<char, TrieNode*> children;
    // Optimization: Keeping track of word within trie would speed up the recursion
    // since we wont need to pass the cur string as a parameter.
    string word = "";
};

class Trie{
    TrieNode* root;

    public:
    Trie(){
        root = new TrieNode();
    }

    TrieNode* getRootNode(){
        return root;
    }

    void insert(string word){
        TrieNode* cur = root;
        for(char c: word){
            if(cur->children.find(c) == cur->children.end())
                cur->children[c] = new TrieNode();
            cur = cur->children[c];
        }
        cur->word = word;
    }
};

class Solution {
    int m, n;
    vector<vector<int>> dirs;

    void traversal(vector<vector<char>>& board, int i, int j, TrieNode* node, vector<string>& res){
        if(i < 0 || i >= m || j < 0 || j >= n || board[i][j] == '#' || node->children.find(board[i][j]) == node->children.end()){
            return;
        }

        char tmp = board[i][j];
        board[i][j] = '#';
        TrieNode* parent = node;
        node = node->children[tmp];

        if(node->word != ""){
            res.push_back(node->word);
            node->word = "";
        }

        for(auto dir: dirs)
            traversal(board, i + dir[0], j + dir[1], node, res);

        // End of EXPLORATION, restore the original letter in the board.
        board[i][j] = tmp;

        // Optimization: incrementally remove the leaf nodes
        if(node->children.empty())
            parent->children.erase(tmp);
    }
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        m = board.size();
        n = board[0].size();
        dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        
        Trie* trie = new Trie();
        for(string word: words)
            trie->insert(word);

        vector<string> res;
        TrieNode* node = trie->getRootNode();
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                traversal(board, i, j, node, res);
            }
        }
        return res;
    }
};
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