Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).<\/p>\n
For example, this binary tree\u00a0 But the following\u00a0 Note:<\/b> <\/p>\n Python<\/strong><\/p>\n[1,2,2,3,4,4,3]<\/code>\u00a0is symmetric:<\/p>\n 1\r\n \/ \\\r\n 2 2\r\n \/ \\ \/ \\\r\n3 4 4 3\r\n<\/pre>\n
[1,2,2,null,3,null,3]<\/code>\u00a0is not:<\/p>\n 1\r\n \/ \\\r\n 2 2\r\n \\ \\\r\n 3 3\r\n<\/pre>\n
\nBonus points if you could solve it both recursively and iteratively.<\/p>\n \r\n# Definition for a binary tree node.\r\n# class TreeNode(object):\r\n# def __init__(self, x):\r\n# self.val = x\r\n# self.left = None\r\n# self.right = None\r\n\r\nclass Solution(object):\r\n def isSymmetric(self, root):\r\n """\r\n :type root: TreeNode\r\n :rtype: bool\r\n """\r\n return self.isMirror(root, root)\r\n\r\n def isMirror(self, t1, t2):\r\n if t1 is None and t2 is None:\r\n return True\r\n elif t1 is None or t2 is None:\r\n return False\r\n return t1.val == t2.val and self.isMirror(t1.right, t2.left) and self.isMirror(t1.left, t2.right)\r\n<\/pre>\n","protected":false},"excerpt":{"rendered":"