**Flatten Binary Tree to Linked List LeetCode Solution** says that – Given the `root`

of a binary tree, flatten the tree into a “linked list”:

- The “linked list” should use the same
`TreeNode`

class where the`right`

child pointer points to the next node in the list and the`left`

child pointer is always`null`

. - The “linked list” should be in the same order as a
**pre-order****traversal**of the binary tree.

**Example 1:**

**Input:**

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

**Output:**

[1,null,2,null,3,null,4,null,5,null,6]

**Example 2:**

**Input:**

root = []

**Output:**

[]

**Example 3:**

**Input:**

root = [0]

**Output:**

[0]

**ALGORITHM –**

IDEA –

- In order to flatten a binary tree, we will first find the rightmost element of the left subtree and after we got the rightmost element we will link the right-pointer of that node with a right subtree of a given tree.
- In step 2 we will link the right pointer of the root node with the left-subtree and set the left-subtree as null.
- In step 3 now our root node is a right-subtree node same process will happen with this node and the process will still continue until all the left parts become null.

**Approach for Flatten Binary Tree to Linked List Leetcode Solution –**

– At first, i will run a loop i.e. while(root != null) then will take two variables and store the left-subtree.

– then will check check for the rightmost node of left-subtree by using while(k.left != null) and will link that node with right subtree using (k.right = root.right).

– then link right pointer of root node with left subtree(root.right = left) and set left pointer of root node as null(root.left=null) and will update by ( root = root.right ) so now root is right subtree node.

– this process will continue until all left-subtree parts become right subtree. Hence, the binary tree will get flattened.

## Python Solution:

class Solution: def flatten(self, root: Optional[TreeNode]) -> None: while(root): if root.left: k = root.left temp = root.left while(k.right): k = k.right k.right = root.right root.right = temp root.left = None root = root.right

## Java Solution:

class Solution { public void flatten(TreeNode root) { while (root != null) { if (root.left != null) { TreeNode k = root.left; TreeNode temp = root.left; while (k.right != null) k = k.right; k.right = root.right; root.right = temp; root.left = null; } root = root.right; } } }

**Time complexity: O(N)**

**Space Complexity: O(1)**

As we have traversed only once, time complexity will be o(n).

and as we haven’t taken any extra space, space complexity will be o(1) constant extra space.

Similar Question – https://www.tutorialcup.com/interview/linked-list/flattening-linked-list.htm