[Python手撕]把二叉树展开为链表

# 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: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """

        cur = root
        while cur:
	# 如果有左子树，争取把左子树全部挪到右子树上
            if cur.left:
                t = cur.left
                while t.right:
                    t = t.right
                t.right = cur.right
                cur.right = cur.left
                cur.left = None
            cur = cur.right