101. 对称二叉树

给你一个二叉树的根节点 root ， 检查它是否轴对称。

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root.left==None and root.right==None:
            return True
        if root.left == None and root.right!=None:
            return False
        if root.right == None and root.left!=None:
            return False

        return self.isSubSymmetric(root.left, root.right)


    def isSubSymmetric(self, lc, rc):
        if lc==None and rc==None:
            return True
        if lc==None and rc!=None or rc==None and lc!=None:
            return False
        # if lc.left==None and rc.right!=None or lc.right==None and rc.left!=None:
        #     return False
        # if lc.right==None and rc.left!=None:
        #     return False
        # if lc.left==None and lc.right==None and rc.left==None and rc.right==None and lc.val!=rc.val:
        #     return False
        # if lc.left==None and lc.right==None and rc.left==None and rc.right==None and lc.val==rc.val:
        #     return True
        #以上判断部分均可在下一次的子节点中进行判断，不需要
        if lc.val ==rc.val:
            return self.isSubSymmetric(lc.left, rc.right) and self.isSubSymmetric(lc.right, rc.left)
        else:
            return False


t1 = TreeNode(2)
t2 = TreeNode(97)
t3 = TreeNode(97)
t4 = TreeNode(47)
t5 = TreeNode(80)
t6 = TreeNode(-7)
t7 = TreeNode(-7)

t1.left = t2
t1.right = t3
t2.right = t4
t3.left = t5
t4.left = t6
t5.right = t7

solution = Solution()
print(solution.isSymmetric(t1))