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Create a sorted linked list from the given Binary Tree

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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

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 <iostream>
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

Javascript




<script>
 
// JavaScript implementation of the approach
 
// A linked list node
class Node {
 
    constructor(data)
    {
        this.data = data;
        this.next = null;
    }
};
 
// A binary tree node
class treeNode {
 
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Function to print the linked list
function print(head)
{
    if (head == null) {
        return;
    }
    var temp = head;
    while (temp != null) {
        document.write( temp.data + " ");
        temp = temp.next;
    }
}
 
// Function to create Linked list from given binary tree
function sortedList(head, 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);
    var newNode = new Node(root.data);
    var temp = head;
    var 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
 
    /* Tree:
         10
        /  \
      15    2
     /  \
    1    5
*/
var 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);
var head = sortedList(null, root);
print(head);
 
 
</script>

Output

1 2 5 10 15 

Time Complexity: O(n2)

Auxiliary Space: O(n)

Another Approach(Using extra space):
Follow the below steps to solve the problem:
1) Create an array to store the all elements of given binary tree.
2) Sort the given array in O(NlogN) time and then traverse the sorted array.
3) While traversing the sorted array then create given linked list for each element.
4) print the created sorted linked list.

Below is the implementation of above approach:

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// a linked list node
struct LNode{
    int data;
    LNode* next;
    LNode(int data){
        this->data = data;
        this->next = NULL;
    }
};
 
// a binary tree node
struct TNode{
    int data;
    TNode* left;
    TNode* right;
    TNode(int data){
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// function to print the linked list
void printList(LNode* head){
    if(head == NULL) return;
    LNode* temp = head;
    while(temp != NULL){
        cout<<temp->data<<" ";
        temp = temp->next;
    }
    cout<<endl;
}
 
// function to store in Inorder fashion
void inOrder(TNode* root, vector<int> &vec){
    if(root == NULL) return;
    inOrder(root->left, vec);
    vec.push_back(root->data);
    inOrder(root->right, vec);
}
 
// function to create sorted linked list from given binary tree
LNode* sortedList(TNode* root){
    // initializing vector to store the elements
    vector<int> vec;
    inOrder(root, vec);
    sort(vec.begin(), vec.end());
    LNode* head = new LNode(-1);
    LNode* temp = head;
    for(int i : vec){
        temp->next = new LNode(i);
        temp = temp->next;
    }
    head = head->next;
    return head;
}
 
// driver code to test above functions
int main(){
    /* Tree:
         10
        /  \
      15    2
     /  \
    1    5
    */
    TNode* root = new TNode(10);
    root->left = new TNode(15);
    root->right = new TNode(2);
    root->left->left = new TNode(1);
    root->left->right = new TNode(5);
     
    LNode* head = sortedList(root);
    printList(head);
    return 0;
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

Java




import java.util.*;
 
// a linked list node
class LNode{
    int data;
    LNode next;
    LNode(int data){
        this.data = data;
        this.next = null;
    }
}
 
// a binary tree node
class TNode{
    int data;
    TNode left;
    TNode right;
    TNode(int data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// main class
class Main{
    // function to print the linked list
    static void printList(LNode head){
        if(head == null) return;
        LNode temp = head;
        while(temp != null){
            System.out.print(temp.data+" ");
            temp = temp.next;
        }
        System.out.println();
    }
 
    // function to store in Inorder fashion
    static void inOrder(TNode root, ArrayList<Integer> vec){
        if(root == null) return;
        inOrder(root.left, vec);
        vec.add(root.data);
        inOrder(root.right, vec);
    }
 
    // function to create sorted linked list from given binary tree
    static LNode sortedList(TNode root){
        // initializing vector to store the elements
        ArrayList<Integer> vec = new ArrayList<Integer>();
        inOrder(root, vec);
        Collections.sort(vec);
        LNode head = new LNode(-1);
        LNode temp = head;
        for(int i : vec){
            temp.next = new LNode(i);
            temp = temp.next;
        }
        head = head.next;
        return head;
    }
 
    // driver code to test above functions
    public static void main(String[] args){
        /* Tree:
             10
            /  \
          15    2
         /  \
        1    5
        */
        TNode root = new TNode(10);
        root.left = new TNode(15);
        root.right = new TNode(2);
        root.left.left = new TNode(1);
        root.left.right = new TNode(5);
         
        LNode head = sortedList(root);
        printList(head);
    }
}

Python




# Python program for the above approach
# a linked list node
class LNode:
    def __init__(self, data):
        self.data = data
        self.next = None
     
 
# a binary tree node
class TNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
     
 
# function to print the linked list
def printList(head):
    if(head is None):
        return
    temp = head
    while(temp is not None):
        print(temp.data)
        temp = temp.next
     
     
# function to store in Inorder fashion
def inOrder(root, vec):
    if(root is None):
        return
    inOrder(root.left, vec)
    vec.append(root.data)
    inOrder(root.right, vec)
     
 
# function to create sorted linked list from given binary tree
def sortedList(root):
    # initialize vector to store the elements
    vec = []
    inOrder(root, vec)
    vec.sort()
    head = LNode(-1)
    temp = head
    for i in vec:
        temp.next = LNode(i)
        temp = temp.next
    head = head.next
    return head
     
 
# driver code to test above function
root = TNode(10)
root.left = TNode(15)
root.right = TNode(2)
root.left.left = TNode(1)
root.left.right = TNode(5)
 
head = sortedList(root)
printList(head)

C#




// C# Program for the above approach
using System;
using System.Collections.Generic;
 
// a linked list node
public class LNode{
    public int data;
    public LNode next;
    public LNode(int item){
        data = item;
        next = null;
    }
}
 
// a binary tree node
public class TNode{
    public int data;
    public TNode left, right;
    public TNode(int item){
        data = item;
        left = right = null;
    }
}
 
class GFG{
    // function to print the linked list
    static void printList(LNode head){
        if(head == null) return;
        LNode temp = head;
        while(temp != null){
            Console.Write(temp.data + " ");
            temp = temp.next;
        }
    }
     
    // function to store in Inorder fashion
    static void inOrder(TNode root, List<int> vec){
        if(root == null) return;
        inOrder(root.left, vec);
        vec.Add(root.data);
        inOrder(root.right, vec);
    }
     
    // function to create sorted linked list from given binary tree
    static LNode sortedList(TNode root){
        // initializing vector to store the elements
        List<int> vec = new List<int>();
        inOrder(root, vec);
        vec.Sort();
        LNode head = new LNode(-1);
        LNode temp = head;
        for(int i = 0; i<vec.Count; i++){
            temp.next = new LNode(vec[i]);
            temp = temp.next;
        }
        head = head.next;
        return head;
    }
     
    // driver code to test above function
    public static void Main(String[] args){
        TNode root = new TNode(10);
        root.left = new TNode(15);
        root.right = new TNode(2);
        root.left.left = new TNode(1);
        root.left.right = new TNode(5);
         
        LNode head = sortedList(root);
        printList(head);
    }
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

Javascript




// JavaScript program for the above approach
// a linked list node
class LNode{
    constructor(data){
        this.data = data;
        this.next = null;
    }
}
 
 
// a binary tree node
class TNode{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// function to print the linked list
function printList(head){
    if(head == null) return;
    let temp = head;
    while(temp != null){
        console.log(temp.data + " ");
        temp = temp.next;
    }
    console.log(" ");
}
 
// function to store the inorder fashion
function inOrder(root, vec){
    if(root == null) return;
    inOrder(root.left, vec);
    vec.push(root.data);
    inOrder(root.right, vec);
}
 
// function to create sorted linked list from given binary tree
function sortedList(root){
    // initializing vector to store the elements
    let vec = [];
    inOrder(root, vec);
    vec.sort(function(a,b){return a-b});
    let head = new LNode(-1);
    let temp = head;
    for(let i = 0; i<vec.length; i++){
        temp.next = new LNode(vec[i]);
        temp = temp.next;
    }
    head = head.next;
    return head;
}
 
// driver program to test above functions
/* Tree:
     10
    /  \
  15    2
 /  \
1    5
*/
let root = new TNode(10);
root.left = new TNode(15);
root.right = new TNode(2);
root.left.left = new TNode(1);
root.left.right = new TNode(5);
 
let head = sortedList(root);
printList(head);

Output

1 2 5 10 15 

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(N) due to extra space.


Last Updated : 14 Apr, 2023
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