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.
Python
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def isSymmetric(self, root): """ :type root: TreeNode :rtype: bool """ return self.isMirror(root, root) def isMirror(self, t1, t2): if t1 is None and t2 is None: return True elif t1 is None or t2 is None: return False return t1.val == t2.val and self.isMirror(t1.right, t2.left) and self.isMirror(t1.left, t2.right)