Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

``````Example:

Input:
board = [
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
words = ["oath","pea","eat","rain"]

Output: ["eat","oath"]

Note:

All inputs are consist of lowercase letters a-z.
The values of words are distinct.
``````

Solution:

``````type Trie struct {
childs[26] *Trie
Count int
}

func (t *Trie) Insert(word string) {
node  := t
lenW := len(word)
for i:= 0; i < lenW; i++ {
idx := word[i] - 'a'
if node.childs[idx] == nil {
node.childs[idx] =  &Trie{}
}
node = node.childs[idx]
}
node.Count++
}

func (t *Trie)StartWith(prefix string) *Trie {
node  := t
lenP := len(prefix)
for i:= 0; i < lenP;i++ {
idx := prefix[i] - 'a'
if node.childs[idx] == nil {
return nil
}
node = node.childs[idx]
}
return node
}

func findWords(board [][]byte, words []string) []string {
trie := &Trie{}
//build trie tree
for i :=0; i < len(words); i++ {
trie.Insert(words[i])
}

result := map[string]bool{}
//start to bactracing
for x := 0; x < len(board); x++ {
for y := 0; y < len(board[0]); y++ {
backTrace(&board, x, y,  trie, "", &result)
}
}

list := []string{}
for k, _ := range result {
list = append(list, k)
}
return list
}

func backTrace(board *([][]byte),  x, y int, trie *Trie, prefix string, result *map[string]bool) {
lenX := len(*board)
lenY := len((*board)[0])

if x >= 0 && x < lenX && y >= 0 && y < lenY {
char := (*board)[x][y]
if char ==  '#' {
return
}
(*board)[x][y] = '#' //mark visited
newPrefix := prefix + string(char)
node := trie.StartWith(newPrefix)

if node != nil {
backTrace(board , x-1, y, trie, newPrefix, result)
backTrace(board , x+1, y, trie, newPrefix, result)
backTrace(board , x, y -1 , trie, newPrefix, result)
backTrace(board , x, y+1, trie, newPrefix, result)

if node.Count > 0 {
(*result)[newPrefix] = true
}
}
(*board)[x][y] = char

}
}``````