114. 二叉树展开为链表

1 题目

链接：二叉树展开为链表

2 Python

2.1 方法一：递归

思想：左子树接到根节点的右子树位置，右子树接到左子树的右子树位置
从二叉树的左下角往右上角逐个操作

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        # 出口函数
        if not root:
            return None;
        # 递归过程
        self.flatten(root.right);
        self.flatten(root.left);
        if root.left:
            right = root.right;
            root.right = root.left;
            root.left = None;
            # 找到左子树（已经接到了根节点的右子树位置）最下层的右子树
            while root.right:
                root = root.right;
            root.right = right;

来源：CSDN

作者：闲人泽阳

链接：https://blog.csdn.net/weixin_43969686/article/details/104208439