Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
<pre>Input: root = [1,2,2,3,4,4,3] Output: true </pre>
Example 2:
<pre>Input: root = [1,2,2,null,3,null,3] Output: false </pre>
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?</div>
题目大意：
判断二叉树是否对称
解题思路：
Easy题，但难点是转化成比较两棵树是否对称
解题步骤：
N/A
注意事项：
- 难点是转化成比较两棵树是否对称
- 还要比较值相等，root.val == root2.val
Python代码：
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9def isSymmetric(self, root: TreeNode) -> bool:
return self.is_symmetric(root.left, root.right)
def is_symmetric(self, root, root2):
if not root and not root2:
return True
if not root or not root2:
return False
return root.val == root2.val and self.is_symmetric(root.left, root2.right) and self.is_symmetric(root.right, root2.left)
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
