对称二叉树

递归解法：

我的解法是层序遍历然后每层取eq(line，line[::-1]):

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        import operator
        result = []
        self.levelExpler(root, result, 1)
        for line in result:
            if not operator.eq(line, line[::-1]):
                return False
        return True
        
    def levelExpler(self, root, result, level):
        if root is None:
            if len(result) < level:
                result.append([root])
            else:
                result[level - 1].append(root)
            return
        if len(result) < level:
            result.append([root.val])
        else:
            result[level - 1].append(root.val)
        self.levelExpler(root.left, result, level + 1)
        self.levelExpler(root.right, result, level + 1)

大佬解法：这种解法利用了1生2，2生4，。。。这种方法，符合了二叉树的每层大小，利用left.left and right.right和left.right and right.left这种递归方法。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if not root:
            return True
        def mirror(left_node, right_node):
            if not (left_node or right_node): #当左右节点都为空时 
                return True
            elif not (left_node and right_node):
                return False
            if left_node and right_node:
                if left_node.val == right_node.val:
                    return mirror(left_node.left, right_node.right) and mirror(left_node.right, right_node.left)
                else:
                    return False
        return mirror(root.left, root.right)

迭代解法：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        
        if root is None: return True
        
        isSym = True
        L = [root]
        R = [root]
        while isSym and len(L) and len(R):
            l, r = L.pop(), R.pop()
            if l is None and r is None:
                isSym = True
            elif l is None or r is None:
                isSym = False
            elif l.val == r.val:
                L.append(l.left)
                R.append(r.right)
                L.append(l.right)
                R.append(r.left)
            else:
                isSym = False
        return isSym