给定一个二叉树，检查它是否是镜像对称的。

解法1 递归

# 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
        return self.recur(root.left, root.right)
        
    
    def recur(self, a, b):
        
        if a is None and b is not None:
            return False
        if a is not None and b is None:
            return False
        if a is None and b is None:
            return True
        
        r = a.val == b.val
        p = self.recur(a.left, b.right)
        q = self.recur(a.right, b.left)
        
        return r and p and q

解法2 迭代

• 队列

# 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
        # 迭代
        return self.itera(root.left, root.right)
        
        
        
    def itera(self, a, b):
        
        l = [(a, b)]
        
        while l:
            a, b = l.pop(0)
            
            if a and b and a.val == b.val:
                l.append((a.left, b.right))
                l.append((a.right, b.left))
            
            elif not a and not b:
                pass
            else:
                return False
            
            
            # elif not a and b: # a is None and b is not None
            #     return False
            # elif a and not b: # a is not None and b is None
            #     return False
            # elif not a and not b: # a is None and b is None
            #     pass
            # else: # a and b and a.val != b.val
            #     return False 
            
        return True