0x3f 第33天 redis+链表

1.二叉树的中序遍历

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        def dfs(node):
            if node is None:
                return 
            a = dfs(node.left)
            ans.append(node.val)
            b = dfs(node.right)
        dfs(root)
        return ans

2.翻转二叉树

前序的思路

1.先翻转当前状态的左右节点

2.遍历左

3.遍历右

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

3.二叉树的直径