【第三轮】7.14-7.20 CareerCup 4.8

【解题思路】
Recursively check if t2 is the subtree of t1, the check is also a recursive function which checks whether two tree match or not.
【时间复杂度】
O(MN)
【空间复杂度】
O(N)
【gist link】
https://gist.github.com/iwilbert/96e4a0e45b2142f058e0

好久不来，上次说暂时休息后就一下子拉下来了 >.< 继续
【解题思路】
for all nodes in T1 that the same as the root of T2, check if this is the subtree of T1 that the same as T2
【时间复杂度】
O(n+km)
worst case : O(m*n)
ps: n--Node# in t1
     m--Node# in t2
      k--occurence of t2 root in t1
【空间复杂度】
O(log m + log n)  for recursion calls
【gist link】 https://gist.github.com/JoyceeLee/690e6bba1b4631aa9fff