Related Topics: “栈”: https://leetcode.com/tag/stack/ “树”: https://leetcode.com/tag/tree/ “深度优先搜索”: https://leetcode.com/tag/depth-first-search/ “链表”: https://leetcode.com/tag/linked-list/ “二叉树”: https://leetcode.com/tag/binary-tree/ Similar Questions: “扁平化多级双向链表”: https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/

Problem: Link to heading

给你二叉树的根结点 root ，请你将它展开为一个单链表：

  • 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
  • 展开后的单链表应该与二叉树 先序遍历 顺序相同。

示例 1：

输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [0]
输出：[0]

提示：

  • 树中结点数在范围 [0, 2000] 内
  • -100 <= Node.val <= 100

**进阶：**你可以使用原地算法（O(1) 额外空间）展开这棵树吗？

Solution: Link to heading

https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/solution/tong-su-yi-dong-de-si-lu-fen-xi-zong-gon-bi7h/

lass Solution {
public:
    void flatten(TreeNode* root) {
        if(root == nullptr){
            return;
        }
        flatten(root->left);
        flatten(root->right);

        TreeNode* l = root->left;
        TreeNode* r = root->right;

        root->left = nullptr;
        root->right = l;

        while(root->right != nullptr){
            root = root->right;
        }
        root->right = r;
    }
};