Why Am I not getting expected output? - binary-tree

Question

I am a beginner in trees.I defined a findNode function to return the node.Then I recursively check if both node as root and subtree is same.But I have not getting the expected output for the below test case. How can it be altered.
Thanks in advance
Expected output:true
Input: Tree:[1,1]
subTree:[1]
but the output I am getting is false
leetcode problem: https://leetcode.com/problems/subtree-of-another-tree/submissions/
def findNode(self,root,subRoot):
if root==None:
return None
if root.val==subRoot.val:
return root
if root.left:
return self.findNode(root.left,subRoot)
if root.right:
return self.findNode(root.right,subRoot)
def check(self,root,subRoot):
if root==None and subRoot==None:
return True
if root==None or subRoot==None:
return False
if root.val!=subRoot.val:
return False
return self.check(root.right,subRoot.right) and self.check(root.left,subRoot.left)
def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
node=self.findNode(root,subRoot)
return self.check(node,subRoot)
```

Answer 1

There was a problem in your code because once you find a matching node with subRoot node you stop searching and you run check() function, which can return false and your program will return that false without checking deeper subtrees, which can be matched. Here is my code:
class Solution:
def findMatchingTree(self,root,subRoot):
if root==None:
return False
return self.check(root, subRoot) or self.findMatchingTree(root.left,subRoot) or self.findMatchingTree(root.right,subRoot)
def check(self,root,subRoot):
if root==None and subRoot==None:
return True
if root==None or subRoot==None:
return False
if root.val!=subRoot.val:
return False
return self.check(root.right,subRoot.right) and self.check(root.left,subRoot.left)
def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
return self.findMatchingTree(root,subRoot)
If you want to make code faster I suggests you to search for roots from the leafs of tree.