【 二叉树展开为链表】python刷题记录

先序遍历读取简化：

 把左子树变为右子树，把右子树移动到”新右子树“右下角

# 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: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        #判空
        if not root:
            return
        #存右子树
        right=root.right
        #分别递归左右子树
        self.flatten(root.left)
        self.flatten(root.right)
        root.right=root.left
        root.left=None
        #可移动的结点指针
        node=root
        while node.right:
             node=node.right
        #遍历完左子树（"新右子树")，连接原右子树
        node.right=right