<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>

class TrieNode:
    def __init__(self):
        self.children = {}
        self.isEnd = False

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        root = TrieNode()
        
        # Create a Trie Datastructure from the given words
        for word in words:
            node = root
            # Insert the word into the Trie
            for c in word:
                if c not in node.children:
                    node.children[c] = TrieNode()
                node = node.children[c]
            node.isEnd = True

        found = []
        visited = set()

        def dfs(i, j,node,path):
            # If the current node, is at the end of a word.
            if node.isEnd:
                found.append(path)
                node.isEnd = False
                
            # Check if nodes are valid in the next 4 directions
            for move_i, move_j in [(1,0),(0,1),(-1,0),(0,-1)]:
                new_i, new_j = i + move_i , j + move_j 
                if 0 <= new_i < len(board) and 0 <= new_j < len(board[0]):
                    if board[new_i][new_j] in node.children and (new_i,new_j) not in visited:
                        curr_letter = board[new_i][new_j]
                        child = node.children[curr_letter]
                        # Backtrack 
                        visited.add((new_i,new_j))
                        dfs(new_i,new_j,child,path + curr_letter)
                        visited.remove((new_i,new_j))
                        
        # Iterate through the array. If any letter is starting of any word, perform DFs on that.
        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] in root.children:
                    visited.add((i,j))
                    dfs(i,j,root.children[board[i][j]],board[i][j])
                    visited.pop()

        return found



        

            

        
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