# Flatten Binary Tree to Linked List LeetCode Solution

Here, We see Flatten Binary Tree to Linked List LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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]

## Flatten Binary Tree to Linked List Leetcode Solution C++

``````class Solution {
public:
void flatten(TreeNode* root) {
if( root )
{
TreeNode* temp = root->right;
root->right = root->left;
root->left = nullptr;
TreeNode* node = root;
while( node->right )
{
node = node->right;
}
node->right = temp;
flatten( root->right );
}
return;
}
};```Code language: PHP (php)```

## Flatten Binary Tree to Linked List Leetcode Solution Java

``````class Solution {
public void flatten(TreeNode root) {
if (root == null) return;
TreeNode left = root.left;
TreeNode right = root.right;
root.left = null;
flatten(left);
flatten(right);
root.right = left;
TreeNode cur = root;
while (cur.right != null) cur = cur.right;
cur.right = right;
}
}```Code language: JavaScript (javascript)```

## Flatten Binary Tree to Linked List Solution JavaScript

``````var flatten = function(root) {
let curr = root
while (curr) {
if (curr.left) {
let runner = curr.left
while (runner.right) runner = runner.right
runner.right = curr.right, curr.right = curr.left, curr.left = null
}
curr = curr.right
}
};```Code language: JavaScript (javascript)```

## Flatten Binary Tree to Linked List Solution Python

``````class Solution(object):
def flatten(self, root):
curr = root
while curr:
if curr.left:
runner = curr.left
while runner.right: runner = runner.right
runner.right, curr.right, curr.left = curr.right, curr.left, None
curr = curr.right``````
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