<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 Node {
  unordered_map<char, Node*> children;
  string word;
};

class Solution {
private:
  vector<vector<char>> brd;
  vector<string> res;
public:
  void Search(int row, int col, Node* parent) {
    char letter = brd[row][col];
    // we can always get a valid node because the Search() has a condition check
    Node* node = parent->children[letter];

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

    brd[row][col] = '#';

    vector<int> row_off = { -1, 0, 1, 0 };
    vector<int> col_off = { 0, -1, 0, 1 };
    for (int i = 0; i < 4; ++i) {
      int r = row + row_off[i];
      int c = col + col_off[i];
      if (r < 0 || r >= brd.size() || c < 0 || c >= brd[0].size()) {
        continue;
      }
      if (node->children.find(brd[r][c]) != node->children.end()) {
        Search(r, c, node);
      }
    }

    brd[row][col] = letter;
    if (node->children.empty()) {
      parent->children.erase(letter);
    }
  }

  vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
    Node* root = new Node();
    for (auto word : words) {
      Node* node = root;
      for (char c : word) {
        if (node->children.find(c) != node->children.end()) {
          node = node->children[c];
        } else {
          Node* node2 = new Node();
          node->children[c] = node2;
          node = node2;
        }
      }
      node->word = word;
    }

    brd = board;
    for (int r = 0; r < brd.size(); ++r) {
      for (int c = 0; c < brd[0].size(); ++c) {
        // scan the whole board; if any letter matches Trie starting letter,
        // start the recursive search
        if (root->children.find(brd[r][c]) != root->children.end()) {
          Search(r, c, root);
        }
      }
    }

    return res;
  }
};
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