# Flatten Binary Tree to Linked List LeetCode Solution Share:

Flatten Binary Tree to Linked List 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 = 
Output: 
```

Constraints:

• The number of nodes in the tree is in the range `[0, 2000]`.
• `-100 <= Node.val <= 100`

Time: O(n)
Space: O(h)

### C++

``````class Solution {
public:
void flatten(TreeNode* root) {
if (!root)
return;

flatten(root->left);
flatten(root->right);

TreeNode* const left = root->left;    // flattened left
TreeNode* const right = root->right;  // flattened right

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

// connect the original right subtree
// to the end of new right subtree
TreeNode* rightmost = root;
while (rightmost->right)
rightmost = rightmost->right;
rightmost->right = right;
}
};
``````

### Java

``````
class Solution {
public void flatten(TreeNode root) {
if (root == null)
return;

flatten(root.left);
flatten(root.right);

TreeNode left = root.left;   // flattened left
TreeNode right = root.right; // flattened right

root.left = null;
root.right = left;

// connect the original right subtree
// to the end of new right subtree
TreeNode rightmost = root;
while (rightmost.right != null)
rightmost = rightmost.right;
rightmost.right = right;
}
}
``````

### Python

``````class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
if not root:
return

self.flatten(root.left)
self.flatten(root.right)

left = root.left  # flattened left
right = root.right  # flattened right

root.left = None
root.right = left

# connect the original right subtree
# to the end of new right subtree
rightmost = root
while rightmost.right:
rightmost = rightmost.right
rightmost.right = right
``````

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