My solution to wordsearch 2 on LC: https://leetcode.com/problems/word-search-ii/ a new test case was added that causes TLE for python unless words are removed from the trie which I show here.

wordsearch2.py

class TrieNode:
    def __init__(self):
        self.children = {}
        self.ref = 0 # number of references (ie. size of each map)
        self.end = False
    
    def add_word(self, word):
        node = self
        node.ref += 1
        for c in word:
            if c not in node.children:
                node.children[c] = TrieNode()
            node = node.children[c]
            node.ref += 1
        node.end = True
    
    def remove_word(self, word):
        node = self
        node.ref -= 1
        for c in word:
            # note: could add early exit here is node.ref == 0
            node = node.children[c]
            node.ref -= 1
        node.end = False # ex. ABC, ABC, ABCD don't want to add ABC twice

class Solution:

    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        res = []
        trie = TrieNode()
        for word in words:
            trie.add_word(word)
        
        def dfs(r, c, visited, t, s):
            if (r, c) in visited:
                return
            if r < 0 or r == len(board) or c < 0 or c == len(board[0]):
                return
            curr = board[r][c] # current character
            if curr not in t.children:
                return
            if t.children[curr].ref <= 0:
                return
            
            visited.add((r,c)) # add to visited set down here or we'd have to remove it in earlier returns too
            if t.children[curr].end: # word found
                res.append(s+curr)
                trie.remove_word(s+curr) # remove from root (prevents TLE on python solution)

            dfs(r+1, c, visited, t.children[curr], s+curr)
            dfs(r-1, c, visited, t.children[curr], s+curr)
            dfs(r, c+1, visited, t.children[curr], s+curr)
            dfs(r, c-1, visited, t.children[curr], s+curr)
            visited.remove((r, c)) # backtrack! (undo visited path)
            
        for r in range(len(board)):
            for c in range(len(board[0])):
                dfs(r,c,set(),trie,"")    
        return res