101. 对称二叉树

思路：递归。
终止条件是两个节点都为空,return True；
或者两个节点中有一个为空,return False；
或者两个节点的值不相等,return False；

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        return self.digui(root.left, root.right)

    def digui(self, left, right):
        if left.val != right.val:
            return False
        if not left or not right:
            return False
        if not left and not right:
            return True
        return self.digui(left.left, right.right) and self.digui(left.right, right.left)