Word Ladder
For this problem, BFS is better than DFS.
DFS, it's like from the first char to the last char of the word, try to match the dictionary. However, if there is not match, then start from last matching char and do the same again. So it is not efficiency.
Using BFS, it is like Level traverse a tree. So it would reduce the times for matching.
Here is the code. Using two queue( the distance Queue is not required. it just used for store the distance every time generate.) to do the BFS. (when mention the BFS, the first thought must be Queue. LOL).
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