【leetcode】114: 二叉树展开为链表

# 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 flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        cur=root
        def dfs(root,ls):
            if root==None:
                return None
            ls.append(root.val) 
            dfs(root.left,ls)

            dfs(root.right,ls)
        ls=[]
        dfs(root,ls)
        for i in ls:
            print(i)

        i=0
        while i<len(ls):

            if i!=0:
                root.right=TreeNode(ls[i])
                root.left=None
                root=root.right
            if i==0:
                root.left==None
            i+=1