思路也很简单,就是后续遍历的同时把所有的左右节点换一下位置就 ok. 代码如下:

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class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return
        left = self.invertTree(root.left)
        right = self.invertTree(root.right)
        root.left = right
        root.right = left
        return root
快来参与讨论吧