Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/
2 2
/ /
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/
2 2
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
class Solution:
def isSymmetric(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
def com(left,right):
if left is None and right is None:
return True
if left is None or right is None:
return False
if left.val!=right.val:
return False
return com(left.left,right.right) and com(left.right,right.left)
if not root:
return True
return com(root.left,root.right)