题目如下：
Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.
Example 1:
Given tree s:3
/ \
4 5
/ \
1 2Given tree t:
4
/ \
1 2Return true, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:3
/ \
4 5
/ \
1 2
/
0Given tree t:
4
/ \
1 2Return false.
解题思路：我的方法很简单，用先序遍历的方法分别遍历这两棵树，并记录起遍历的路径，最后判断t的遍历路径是否是s的遍历路径的子串即可。
代码如下：
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None class Solution(object):
path = ''
def traverse(self,node,direction):
if node == None:
return
self.path += (direction + 'V' + str(node.val)) #节点值加上V前缀
if node.left != None:
self.traverse(node.left,'L') #左右节点分别加上标识
else:
self.path += 'LN'
if node.right != None:
self.traverse(node.right,'R')
else:
self.path += 'RN'
def isSubtree(self, s, t):
"""
:type s: TreeNode
:type t: TreeNode
:rtype: bool
"""
self.path = ''
self.traverse(s,'')
self.path += '#'
self.traverse(t,'')
pl = self.path.split('#')
#print self.path
return pl[0].find(pl[1]) != -1