题目描述

分析

基于二叉树的后序遍历操作

代码

class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if root:
            self.flatten(root.left)
            self.flatten(root.right)
            left=root.left
            right=root.right
            root.left=None
            root.right=left
            p=root
            while p.right!=None:
                p=p.right
            p.right=right