可以把树看成两个树. 根节点不用管. 直接同时遍历根节点的左子树和右子树,然后看左子树的左边等不等于右子树的右边, 以此类推. 其实就是简单的后续遍历,在遍历的时候看看左子树是否等于右子树. 代码如下:
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| class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
if root is None:
return True
def travel(node1, node2):
if node1 is None and node2 is None:
return True
if node1 is None and node2 is not None:
return False
if node1 is not None and node2 is None:
return False
if node1.val != node2.val:
return False
return travel(node1.left, node2.right) and travel(node1.right, node2.left)
return travel(root.left, root.right)
|