# Create a sorted linked list from the given Binary Tree

Given a binary tree, the task is to convert it into a sorted linked list.

Examples:

```Input:
1
/  \
2    3
Output: 1 2 3

Input:
2
/   \
4     8
/  \   / \
7   3  5   1
Output: 1 2 3 4 5 7 8

Input:
3
/
4
/
1
/
9
Output: 1 3 4 9
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Recursively iterate the given binary tree and add each node to its correct position in the resultant linked list (initially empty) using insertion sort.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// A linked list node ` `class` `Node { ` `public``: ` `    ``int` `data; ` `    ``Node* next; ` `    ``Node(``int` `data) ` `    ``{ ` `        ``this``->data = data; ` `        ``this``->next = NULL; ` `    ``} ` `}; ` ` `  `// A binary tree node ` `class` `treeNode { ` `public``: ` `    ``int` `data; ` `    ``treeNode* left; ` `    ``treeNode* right; ` `    ``treeNode(``int` `data) ` `    ``{ ` `        ``this``->data = data; ` `        ``this``->left = NULL; ` `        ``this``->right = NULL; ` `    ``} ` `}; ` ` `  `// Function to print the linked list ` `void` `print(Node* head) ` `{ ` `    ``if` `(head == NULL) { ` `        ``return``; ` `    ``} ` `    ``Node* temp = head; ` `    ``while` `(temp != NULL) { ` `        ``cout << temp->data << ``" "``; ` `        ``temp = temp->next; ` `    ``} ` `} ` ` `  `// Function to create Linked list from given binary tree ` `Node* sortedList(Node* head, treeNode* root) ` `{ ` `    ``// return head if root is null ` `    ``if` `(root == NULL) { ` `        ``return` `head; ` `    ``} ` ` `  `    ``// First make the sorted linked list ` `    ``// of the left sub-tree ` `    ``head = sortedList(head, root->left); ` `    ``Node* newNode = ``new` `Node(root->data); ` `    ``Node* temp = head; ` `    ``Node* prev = NULL; ` ` `  `    ``// If linked list is empty add the ` `    ``// node to the head ` `    ``if` `(temp == NULL) { ` `        ``head = newNode; ` `    ``} ` `    ``else` `{ ` ` `  `        ``// Find the correct position of the node ` `        ``// in the given linked list ` `        ``while` `(temp != NULL) { ` `            ``if` `(temp->data > root->data) { ` `                ``break``; ` `            ``} ` `            ``else` `{ ` `                ``prev = temp; ` `                ``temp = temp->next; ` `            ``} ` `        ``} ` ` `  `        ``// Given node is to be attached ` `        ``// at the end of the list ` `        ``if` `(temp == NULL) { ` `            ``prev->next = newNode; ` `        ``} ` `        ``else` `{ ` ` `  `            ``// Given node is to be attached ` `            ``// at the head of the list ` `            ``if` `(prev == NULL) { ` `                ``newNode->next = temp; ` `                ``head = newNode; ` `            ``} ` `            ``else` `{ ` ` `  `                ``// Insertion in between the list ` `                ``newNode->next = temp; ` `                ``prev->next = newNode; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Now add the nodes of the right sub-tree ` `    ``// to the sorted linked list ` `    ``head = sortedList(head, root->right); ` `    ``return` `head; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``/* Tree: ` `         ``10 ` `        ``/  \ ` `      ``15    2 ` `     ``/  \ ` `    ``1    5 ` `*/` `    ``treeNode* root = ``new` `treeNode(10); ` `    ``root->left = ``new` `treeNode(15); ` `    ``root->right = ``new` `treeNode(2); ` `    ``root->left->left = ``new` `treeNode(1); ` `    ``root->left->right = ``new` `treeNode(5); ` ` `  `    ``Node* head = sortedList(NULL, root); ` `    ``print(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// A linked list node ` `static` `class` `Node ` `{ ` `    ``int` `data; ` `    ``Node next; ` `    ``Node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``this``.next = ``null``; ` `    ``} ` `}; ` ` `  `// A binary tree node ` `static` `class` `treeNode ` `{ ` `    ``int` `data; ` `    ``treeNode left; ` `    ``treeNode right; ` `    ``treeNode(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``this``.left = ``null``; ` `        ``this``.right = ``null``; ` `    ``} ` `}; ` ` `  `// Function to print the linked list ` `static` `void` `print(Node head) ` `{ ` `    ``if` `(head == ``null``)  ` `    ``{ ` `        ``return``; ` `    ``} ` `    ``Node temp = head; ` `    ``while` `(temp != ``null``) ` `    ``{ ` `        ``System.out.print(temp.data + ``" "``); ` `        ``temp = temp.next; ` `    ``} ` `} ` ` `  `// Function to create Linked list from given binary tree ` `static` `Node sortedList(Node head, treeNode root) ` `{ ` `    ``// return head if root is null ` `    ``if` `(root == ``null``)  ` `    ``{ ` `        ``return` `head; ` `    ``} ` ` `  `    ``// First make the sorted linked list ` `    ``// of the left sub-tree ` `    ``head = sortedList(head, root.left); ` `    ``Node newNode = ``new` `Node(root.data); ` `    ``Node temp = head; ` `    ``Node prev = ``null``; ` ` `  `    ``// If linked list is empty add the ` `    ``// node to the head ` `    ``if` `(temp == ``null``)  ` `    ``{ ` `        ``head = newNode; ` `    ``} ` `    ``else` `    ``{ ` ` `  `        ``// Find the correct position of the node ` `        ``// in the given linked list ` `        ``while` `(temp != ``null``) ` `        ``{ ` `            ``if` `(temp.data > root.data)  ` `            ``{ ` `                ``break``; ` `            ``} ` `            ``else`  `            ``{ ` `                ``prev = temp; ` `                ``temp = temp.next; ` `            ``} ` `        ``} ` ` `  `        ``// Given node is to be attached ` `        ``// at the end of the list ` `        ``if` `(temp == ``null``) ` `        ``{ ` `            ``prev.next = newNode; ` `        ``} ` `        ``else`  `        ``{ ` ` `  `            ``// Given node is to be attached ` `            ``// at the head of the list ` `            ``if` `(prev == ``null``)  ` `            ``{ ` `                ``newNode.next = temp; ` `                ``head = newNode; ` `            ``} ` `            ``else` `            ``{ ` ` `  `                ``// Insertion in between the list ` `                ``newNode.next = temp; ` `                ``prev.next = newNode; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Now add the nodes of the right sub-tree ` `    ``// to the sorted linked list ` `    ``head = sortedList(head, root.right); ` `    ``return` `head; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``/* Tree: ` `        ``10 ` `        ``/ \ ` `    ``15 2 ` `    ``/ \ ` `    ``1 5 ` `*/` `    ``treeNode root = ``new` `treeNode(``10``); ` `    ``root.left = ``new` `treeNode(``15``); ` `    ``root.right = ``new` `treeNode(``2``); ` `    ``root.left.left = ``new` `treeNode(``1``); ` `    ``root.left.right = ``new` `treeNode(``5``); ` ` `  `    ``Node head = sortedList(``null``, root); ` `    ``print(head); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// A linked list node ` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next; ` `    ``public` `Node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``this``.next = ``null``; ` `    ``} ` `}; ` ` `  `// A binary tree node ` `class` `treeNode ` `{ ` `    ``public` `int` `data; ` `    ``public` `treeNode left; ` `    ``public` `treeNode right; ` `    ``public` `treeNode(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``this``.left = ``null``; ` `        ``this``.right = ``null``; ` `    ``} ` `}; ` ` `  `// Function to print the linked list ` `static` `void` `print(Node head) ` `{ ` `    ``if` `(head == ``null``)  ` `    ``{ ` `        ``return``; ` `    ``} ` `    ``Node temp = head; ` `    ``while` `(temp != ``null``) ` `    ``{ ` `        ``Console.Write(temp.data + ``" "``); ` `        ``temp = temp.next; ` `    ``} ` `} ` ` `  `// Function to create Linked list  ` `// from given binary tree ` `static` `Node sortedList(Node head, treeNode root) ` `{ ` `    ``// return head if root is null ` `    ``if` `(root == ``null``)  ` `    ``{ ` `        ``return` `head; ` `    ``} ` ` `  `    ``// First make the sorted linked list ` `    ``// of the left sub-tree ` `    ``head = sortedList(head, root.left); ` `    ``Node newNode = ``new` `Node(root.data); ` `    ``Node temp = head; ` `    ``Node prev = ``null``; ` ` `  `    ``// If linked list is empty add the ` `    ``// node to the head ` `    ``if` `(temp == ``null``)  ` `    ``{ ` `        ``head = newNode; ` `    ``} ` `    ``else` `    ``{ ` ` `  `        ``// Find the correct position of the node ` `        ``// in the given linked list ` `        ``while` `(temp != ``null``) ` `        ``{ ` `            ``if` `(temp.data > root.data)  ` `            ``{ ` `                ``break``; ` `            ``} ` `            ``else` `            ``{ ` `                ``prev = temp; ` `                ``temp = temp.next; ` `            ``} ` `        ``} ` ` `  `        ``// Given node is to be attached ` `        ``// at the end of the list ` `        ``if` `(temp == ``null``) ` `        ``{ ` `            ``prev.next = newNode; ` `        ``} ` `        ``else` `        ``{ ` ` `  `            ``// Given node is to be attached ` `            ``// at the head of the list ` `            ``if` `(prev == ``null``)  ` `            ``{ ` `                ``newNode.next = temp; ` `                ``head = newNode; ` `            ``} ` `            ``else` `            ``{ ` ` `  `                ``// Insertion in between the list ` `                ``newNode.next = temp; ` `                ``prev.next = newNode; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Now add the nodes of the right sub-tree ` `    ``// to the sorted linked list ` `    ``head = sortedList(head, root.right); ` `    ``return` `head; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``/* Tree: ` `        ``10 ` `        ``/ \ ` `    ``15 2 ` `    ``/ \ ` `    ``1 5 ` `    ``*/` `    ``treeNode root = ``new` `treeNode(10); ` `    ``root.left = ``new` `treeNode(15); ` `    ``root.right = ``new` `treeNode(2); ` `    ``root.left.left = ``new` `treeNode(1); ` `    ``root.left.right = ``new` `treeNode(5); ` ` `  `    ``Node head = sortedList(``null``, root); ` `    ``print(head); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# A linked list node ` `class` `Node: ` `    ``def` `__init__(``self``, data ``=` `0``): ` `        ``self``.data ``=` `data ` `        ``self``.``next` `=` `None` ` `  `# A binary tree node ` `class` `treeNode: ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `     `  `# Function to print the linked list ` `def` `print_(head): ` ` `  `    ``if` `(head ``=``=` `None``):  ` `        ``return` ` `  `    ``temp ``=` `head ` `    ``while` `(temp !``=` `None``): ` `        ``print` `( temp.data, end ``=` `" "` `) ` `        ``temp ``=` `temp.``next` `     `  `# Function to create Linked list from given binary tree ` `def` `sortedList( head, root): ` ` `  `    ``# return head if root is None ` `    ``if` `(root ``=``=` `None``) : ` `     `  `        ``return` `head ` ` `  `    ``# First make the sorted linked list ` `    ``# of the left sub-tree ` `    ``head ``=` `sortedList(head, root.left) ` `    ``newNode ``=` `Node(root.data) ` `    ``temp ``=` `head ` `    ``prev ``=` `None` ` `  `    ``# If linked list is empty add the ` `    ``# node to the head ` `    ``if` `(temp ``=``=` `None``) : ` `        ``head ``=` `newNode ` `     `  `    ``else``: ` ` `  `        ``# Find the correct position of the node ` `        ``# in the given linked list ` `        ``while` `(temp !``=` `None``): ` `         `  `            ``if` `(temp.data > root.data) : ` `                ``break` `             `  `            ``else``: ` `                ``prev ``=` `temp ` `                ``temp ``=` `temp.``next` ` `  `        ``# Given node is to be attached ` `        ``# at the end of the list ` `        ``if` `(temp ``=``=` `None``): ` `            ``prev.``next` `=` `newNode ` `         `  `        ``else``: ` `             `  `            ``# Given node is to be attached ` `            ``# at the head of the list ` `            ``if` `(prev ``=``=` `None``) : ` `                ``newNode.``next` `=` `temp ` `                ``head ``=` `newNode ` `             `  `            ``else``: ` ` `  `                ``# Insertion in between the list ` `                ``newNode.``next` `=` `temp ` `                ``prev.``next` `=` `newNode ` `             `  `    ``# Now add the nodes of the right sub-tree ` `    ``# to the sorted linked list ` `    ``head ``=` `sortedList(head, root.right) ` `    ``return` `head ` ` `  `# Driver code ` ` `  `# Tree: ` `# 10 ` `# / \ ` `# 15 2 ` `# / \ ` `#1 5 ` ` `  `root ``=` `treeNode(``10``) ` `root.left ``=` `treeNode(``15``) ` `root.right ``=` `treeNode(``2``) ` `root.left.left ``=` `treeNode(``1``) ` `root.left.right ``=` `treeNode(``5``) ` ` `  `head ``=` `sortedList(``None``, root) ` ` `  `print_(head) ` ` `  `# This code is contributed by Arnab Kundu `

Output:

```1 2 5 10 15
```

Time Complexity: O(n2)

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