Update 0126._Word_Ladder_II.md

docs/Leetcode_Solutions/Python/0126._Word_Ladder_II.md

-### 126. Word Ladder II
+# 126. Word Ladder II
 
-题目:
+**<font color=red>难度: Hard</font>**
 
-<https://leetcode.com/problems/word-ladder-ii/>
+## 刷题内容
 
-难度:
+> 原题连接
 
-Hard
+* https://leetcode.com/problems/word-ladder-ii/
+
+> 内容描述
+
+```
+Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:
+
+Only one letter can be changed at a time
+Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
+Note:
+
+Return an empty list if there is no such transformation sequence.
+All words have the same length.
+All words contain only lowercase alphabetic characters.
+You may assume no duplicates in the word list.
+You may assume beginWord and endWord are non-empty and are not the same.
+Example 1:
+
+Input:
+beginWord = "hit",
+endWord = "cog",
+wordList = ["hot","dot","dog","lot","log","cog"]
+
+Output:
+[
+  ["hit","hot","dot","dog","cog"],
+  ["hit","hot","lot","log","cog"]
+]
+Example 2:
+
+Input:
+beginWord = "hit"
+endWord = "cog"
+wordList = ["hot","dot","dog","lot","log"]
+
+Output: []
+
+Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(len(word) * len(dictionary))******- 空间复杂度: O(len(word))******
 
 其实关键在于怎么优化和表示图
 
@@ -43,7 +86,8 @@ Hard
 
 routine 字典，然后再根据这个来寻找路径
 
-`{'cog': ['log', 'dog'], 'hit': [], 'log': ['lot'], 'dog': ['dot'], 'hot': ['hit'], 'lot': ['hot'], 'dot': ['hot']}`
+
+```{'cog': ['log', 'dog'], 'hit': [], 'log': ['lot'], 'dog': ['dot'], 'hot': ['hit'], 'lot': ['hot'], 'dot': ['hot']}```
 
 ```'cog': ['log', 'dog']```这里的意思就是说在走到```'cog'```之前尝试过了```'log'```和```'dog'```，即previous tried node
 
@@ -60,7 +104,6 @@ class Solution(object):
         :type wordList: List[str]
         :rtype: List[List[str]]
         """
-
         def backtrack(result, trace, path, word):
             if len(trace[word]) == 0:
                 result.append([word] + path)