【leetcode】剑指 Offer 28. 对称的二叉树（python）（101. 对称二叉树，这两个题完全一样）递归 + 迭代（队列）

方法一：递归

# 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 recur(self, L, R):
        if (not L) and (not R):
            return True
        if (not L) or (not R) or L.val != R.val:
            return False   
        return self.recur(L.left, R.right) and self.recur(L.right, R.left) 

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

方法二：迭代（队列）

【引用学习大佬解法，动画解释请移步大佬题解】，本文仅take notes

# 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):
        if not root :
            return True
        if (not(root.left) and root.right) or (not(root.right) and root.left):
            return False
        que = [root.left, root.right]
        while que:
            L = que.pop(0)
            R = que.pop(0)
            if (not L) and (not R):
                continue
            if (not L) or (not R) or L.val != R.val:
                return False
            que.append(L.left)
            que.append(R.right)

            que.append(L.right)
            que.append(R.left)
        return True