E

Not Done

```

/*

Flatten Binary Tree to Linked List

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the next pointer in ListNode.

Example

1

\

1 2

/ \ \

2 5 => 3

/ \ \ \

3 4 6 4

\

5

\

6

Note

Don't forget to mark the left child of each node to null.

Or you will get Time Limit Exceeded or Memory Limit Exceeded.

Challenge

Do it in-place without any extra memory.

Tags Expand

Binary Tree Depth First Search

*/

/**

* Definition of TreeNode:

* public class TreeNode {

* public int val;

* public TreeNode left, right;

* public TreeNode(int val) {

* this.val = val;

* this.left = this.right = null;

* }

* }

*/

public class Solution {

/**

* @param root: a TreeNode, the root of the binary tree

* @return: nothing

*/

public TreeNode parentNode = null;

public void flatten(TreeNode root) {

if (root == null) {

return;

}

if (parentNode != null) {

parentNode.left = null;

parentNode.right = root;

}

parentNode = root;

TreeNode right = root.right;

flatten(root.left);

flatten(right);

}

}

```