<p>Given an <code>m x n</code> grid of characters <code>board</code> and a string <code>word</code>, return <code>true</code> <em>if</em> <code>word</code> <em>exists in the grid</em>.</p>

<p>The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCCED&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;SEE&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/word3.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCB&quot;
<strong>Output:</strong> false
</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;= 6</code></li>
	<li><code>1 &lt;= word.length &lt;= 15</code></li>
	<li><code>board</code> and <code>word</code> consists of only lowercase and uppercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you use search pruning to make your solution faster with a larger <code>board</code>?</p>

class Solution {
    struct T{
        T * ch[60];
        T() {
            for(int i=0; i<60; i++) ch[i] = nullptr;
        }
    };
    bool f[6][6];
    void ff(T * rt, int i, int j, int c, int tl, vector<vector<char>>& board){
        if(c > tl) return;
        f[i][j]=1;
        if((i-1)>=0 && f[i-1][j] == 0){
            if(rt->ch[board[i-1][j]-'A'] != nullptr){
                ff(rt->ch[board[i-1][j]-'A'], i-1, j, c+1, tl, board);
            } else {
                T * tr = new T();
                rt->ch[board[i-1][j]-'A'] = tr;
                ff(tr, i-1, j, c+1, tl, board);
            }
        }
        if((j-1)>=0 && f[i][j-1] == 0){
            if(rt->ch[board[i][j-1]-'A'] != nullptr){
                ff(rt->ch[board[i][j-1]-'A'], i, j-1, c+1, tl, board);
            } else {
                T * tr = new T();
                rt->ch[board[i][j-1]-'A'] = tr;
                ff(tr, i, j-1, c+1, tl, board);
            }
        }
        if((i+1)<board.size() && f[i+1][j] == 0){
            if(rt->ch[board[i+1][j]-'A'] != nullptr){
                ff(rt->ch[board[i+1][j]-'A'], i+1, j, c+1, tl, board);
            } else {
                T * tr = new T();
                rt->ch[board[i+1][j]-'A'] = tr;
                ff(tr, i+1, j, c+1, tl, board);
            }
        }
        if((j+1)<board[0].size() && f[i][j+1] == 0){
            if(rt->ch[board[i][j+1]-'A'] != nullptr){
                ff(rt->ch[board[i][j+1]-'A'], i, j+1, c+1, tl, board);
            } else {
                T * tr = new T();
                rt->ch[board[i][j+1]-'A'] = tr;
                ff(tr, i, j+1, c+1, tl, board);
            }
        }
        f[i][j] = 0;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {
        T * rt = new T();
        for(int i=0; i<board.size(); i++){
            for(int j=0; j<board[i].size(); j++) if(f[i][j] == 0){
                if(rt->ch[board[i][j]-'A'] != nullptr){
                    ff(rt->ch[board[i][j]-'A'], i, j, 1, word.size(), board);
                } else {
                    T * tr = new T();
                    rt->ch[board[i][j]-'A'] = tr;
                    ff(tr, i, j, 1, word.size(), board);
                }
            }
        }
        for(int i=0; i<word.size(); i++){
            if(rt->ch[word[i]-'A'] == nullptr){
                return false;
            }
            rt = rt->ch[word[i]-'A'];
        }
        return true;
    }
};
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