题目
给你一个二叉树的根节点 root ， 检查它是否轴对称。
代码
方法一：递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def isSymmetric(self, root: Optional[TreeNode]) -> bool:def check(left,right):if left is None and right is None:return Trueif left is None or right is None:return Falseif left.val!=right.val:return Falseout=check(left.left,right.right)inner=check(left.right,right.left)return out and innerreturn check(root.left,root.right) if root else True方法二:队迭代
这里用的两个队分别存放根节点的左右两个子树的节点，当然也可以只使用一个队
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def isSymmetric(self, root: Optional[TreeNode]) -> bool:if root is None:return Truequeue1=deque([root.left])queue2=deque([root.right])while queue1 and queue2:n=len(queue1)for _ in range(n):node1=queue1.popleft()node2=queue2.popleft()if not node1 and not node2:continueif not node1 or not node2:return False if node1.val != node2.val :return False queue1.append(node1.left)queue1.append(node1.right)queue2.append(node2.right)queue2.append(node2.left)return True
只使用一个队的方法：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def isSymmetric(self, root: Optional[TreeNode]) -> bool:if root is None:return Truequeue=deque() queue.append(root.left)queue.append(root.right)while queue:left=queue.popleft()right=queue.popleft()if not left and not right:continueif not left or not right:return Falseif left.val !=right.val:return Falsequeue.append(left.left)queue.append(right.right)queue.append(left.right)queue.append(right.left)return True
