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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？

题解

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public void flatten(TreeNode root) {List<TreeNode> list = new ArrayList<TreeNode>();preorder(root,list);    for(int i=1;i<list.size();i++){TreeNode preNode = list.get(i-1);TreeNode cur = list.get(i);preNode.right = cur;preNode.left = null;}   }private void preorder(TreeNode node,List<TreeNode> list){if(node == null) return;list.add(node);preorder(node.left,list);preorder(node.right,list);}
}

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TTreeNodereeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public void flatten(TreeNode root) {if (root == null)return;if(root.left != null){TreeNode right = root.right;TreeNode leftRightNode = rightNode(root.left);leftRightNode.right = right;root.right = root.left;root.left = null;}flatten(root.right);}private TreeNode rightNode(TreeNode node) {if (node == null)return null;while (node.right != null)node = node.right;return node;}}