修改为 Python3 模板

Author: lihangqi

lcci/04.10.Check SubTree/README.md

@@ -36,14 +36,14 @@
 
 ```python
 class Solution:
-    def checkSubTree(self,t1,t2):
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
         if t1 == None:
             return False
         if t2 == None:
             return True
         return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
 
-    def dfs(self,t1,t2):
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
         if not t1 and t2 :
             return False
         if not t2 and not t1:

lcci/04.10.Check SubTree/README_EN.md

@@ -57,14 +57,14 @@ Find the t2 node in t1 first, then use the depth-first search (DFS) algorithm to
 
 ```python
 class Solution:
-    def checkSubTree(self,t1,t2):
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
         if t1 == None:
             return False
         if t2 == None:
             return True
         return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
 
-    def dfs(self,t1,t2):
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
         if not t1 and t2 :
             return False
         if not t2 and not t1:

lcci/04.10.Check SubTree/Solution.py

@@ -1,12 +1,12 @@
 class Solution:
-    def checkSubTree(self,t1,t2):
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
         if t1 == None:
             return False
         if t2 == None:
             return True
         return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
 
-    def dfs(self,t1,t2):
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
         if not t1 and t2 :
             return False
         if not t2 and not t1: