这是字典树的第3篇算法，力扣链接。

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words， 返回所有二维网格上的单词 。

单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例 1：

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]

 

type Trie struct {children [26]*Trieword     string
}func (t *Trie) Insert(word string) {node := tfor _, ch := range word {ch -= 'a'if node.children[ch] == nil {node.children[ch] = &Trie{}}node = node.children[ch]}node.word = word
}var dirs = []struct{ x, y int }{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}func findWords(board [][]byte, words []string) []string {t := &Trie{}for _, word := range words {t.Insert(word)}m, n := len(board), len(board[0])seen := map[string]bool{}var dfs func(node *Trie, x, y int)dfs = func(node *Trie, x, y int) {ch := board[x][y]node = node.children[ch-'a']if node == nil {return}if node.word != "" {seen[node.word] = true}board[x][y] = '#'for _, d := range dirs {nx, ny := x+d.x, y+d.yif 0 <= nx && nx < m && 0 <= ny && ny < n && board[nx][ny] != '#' {dfs(node, nx, ny)}}board[x][y] = ch}for i, row := range board {for j := range row {dfs(t, i, j)}}ans := make([]string, 0, len(seen))for s := range seen {ans = append(ans, s)}return ans
}