1. Code
class Solution:
def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
if not p and not q:
return True
# if only one node has a child or children
if not p or not q:
return False
if p.val != q.val:
return False
# keep compare the left and right side of each node
return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
2. Result
Runtime : 24 ms(94.54%), Memory usage : 14.1 MB(50.13%)
(Runtime can be different by a system even if it is a same code.)