Leetcode 79 Word Search (JavaScript) 的问题

各位邦友好，敝人想问一下leetcode
https://leetcode.com/problems/word-search/

这是一道典型的DFS题，我用JavaScript写的：

var exist = function(board, word) {    let row = board.length, col = board[0].length;        let dfs = (word, r, c) =>{        if (!word.length){            return true;        }        else if (r>=0 && r<row && c>=0 && c<col && word[0] == board[r][c]){            let tmp = board[r][c];            board[r][c] = '#';            if (dfs(word.substr(1), r+1, c) || dfs(word.substr(1), r-1, c) || dfs(word.substr(1), r, c+1) || word.substr(1), r, c-1){                return true;            }            board[r][c] = tmp;            return false;        }        else {            return false;        }    }        for (let r=0; r<row; r++){        for (let c=0; c<col; c++){            if (dfs(word, r, c)){                return true;            }        }    }    return false;};

但是以下这个TEST CASE无法通过
[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]
"ABCB"

我用Python写，和上面JavaScript的程式相同的逻辑，就通过了

class Solution(object):    def exist(self, board, word):        """        :type board: List[List[str]]        :type word: str        :rtype: bool        """        row = len(board)        col = len(board[0])                def dfs(word, r, c):            if not word:                return True            elif (r>=0 and r<row and c>=0 and c<col and board[r][c] == word[0]):                tmp = board[r][c]                board[r][c] = '#'                if (dfs(word[1:],r+1,c) or dfs(word[1:],r-1,c) or dfs(word[1:],r,c+1) or dfs(word[1:],r,c-1)):                    return True                board[r][c] = tmp            else:                return False                        for r in range(row):            for c in range(col):                if dfs(word, r, c):                    return True        return False

敝人真的不知道自己的JavaScript程式码错在哪，也不懂该从何debug起
不知这里有没有高手愿意提点，谢谢!