https://leetcode.com/problems/invert-binary-tree/
class Solution: def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: if root is None: return None root.left, root.right = root.right, root.left self.invertTree(root.left) self.invertTree(root.right) return root
时间复杂度:O(n) 空间复杂度:O(h)
Last updated 4 months ago