R7-二叉树篇

一眼递归

递归具有相同子问题的特点。

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return None
        left=self.invertTree(root.left)
        right=self.invertTree(root.right)
        root.left=right
        root.right=left
        return root

ps:

今天很顺利地拿下