问题描述

https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/description/

解题思路

这个题目，用一个数组就能很好的解决。但空间复杂度是O(n).

题目中给的进阶要求，是要空间复杂度为O(1)，所以这就要求我们在递归时就要处理掉。

好，首先，我们先做递归函数的定义。即，递归函数返回的就是一个链表的头节点。

所以我们每一层要做的事情就是递归的处理左子树，递归的处理右子树，然后将左子树生成的链表、根节点以及右子树生成的链表合并成一个最终的链表并返回即可。

代码一

# 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.
        """
        if root is None or (root.left is None and root.right is None):
            return
        li = []
        def dfs(root):
            if root is None:
                return
            li.append(root)
            dfs(root.left)
            dfs(root.right)
        dfs(root)
        for i in range(len(li)-1):
            li[i].left = None
            li[i].right = li[i+1]
        li[-1].left, li[-1].right = None, None

代码二

class Solution:
    def flatten(self, root):
        """
        Do not return anything, modify root in-place instead.
        """
        if root is None:
            return
        if root.left is None and root.right is None:
            return root
        right_child = self.flatten(root.right)
        left_child = self.flatten(root.left)
        root.right = left_child
        root.left = None
        tmp_child = root
        while tmp_child.right:
            tmp_child = tmp_child.right
        tmp_child.right = right_child
        return root