Convert a given Binary Tree to Doubly Linked List | Set 4

Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (left most node in BT) must be head node of the DLL.

Below three different solutions have been discussed for this problem.
Convert a given Binary Tree to Doubly Linked List | Set 1
Convert a given Binary Tree to Doubly Linked List | Set 2
Convert a given Binary Tree to Doubly Linked List | Set 3

In the following implementation, we traverse the tree in inorder fashion. We add nodes at the beginning of current linked list and update head of the list using pointer to head pointer. Since we insert at the beginning, we need to process leaves in reverse order. For reverse order, we first traverse the right subtree before the left subtree. i.e. do a reverse inorder traversal.

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// C++ program to convert a given Binary Tree to Doubly Linked List
#include <bits/stdc++.h>
  
// Structure for tree and linked list
struct Node {
    int data;
    Node *left, *right;
};
  
// Utility function for allocating node for Binary
// Tree.
Node* newNode(int data)
{
    Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return node;
}
  
// A simple recursive function to convert a given
// Binary tree to Doubly Linked List
// root    --> Root of Binary Tree
// head --> Pointer to head node of created doubly linked list
void BToDLL(Node* root, Node*& head)
{
    // Base cases
    if (root == NULL)
        return;
  
    // Recursively convert right subtree
    BToDLL(root->right, head);
  
    // insert root into DLL
    root->right = head;
  
    // Change left pointer of previous head
    if (head != NULL)
        head->left = root;
  
    // Change head of Doubly linked list
    head = root;
  
    // Recursively convert left subtree
    BToDLL(root->left, head);
}
  
// Utility function for printing double linked list.
void printList(Node* head)
{
    printf("Extracted Double Linked list is:\n");
    while (head) {
        printf("%d ", head->data);
        head = head->right;
    }
}
  
// Driver program to test above function
int main()
{
    /* Constructing below tree 
            
            / \ 
            3     6 
        / \     \ 
        1 4     8 
        / \     / \ 
        0 2     7 9 */
    Node* root = newNode(5);
    root->left = newNode(3);
    root->right = newNode(6);
    root->left->left = newNode(1);
    root->left->right = newNode(4);
    root->right->right = newNode(8);
    root->left->left->left = newNode(0);
    root->left->left->right = newNode(2);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(9);
  
    Node* head = NULL;
    BToDLL(root, head);
  
    printList(head);
  
    return 0;
}
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// Java program to convert a given Binary Tree to Doubly Linked List
  
/* Structure for tree and Linked List */
class Node {
    int data;
    Node left, right;
  
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
  
class BinaryTree {
    // root    --> Root of Binary Tree
    Node root;
  
    // head --> Pointer to head node of created doubly linked list
    Node head;
  
    // A simple recursive function to convert a given
    // Binary tree to Doubly Linked List
    void BToDLL(Node root)
    {
        // Base cases
        if (root == null)
            return;
  
        // Recursively convert right subtree
        BToDLL(root.right);
  
        // insert root into DLL
        root.right = head;
  
        // Change left pointer of previous head
        if (head != null)
            head.left = root;
  
        // Change head of Doubly linked list
        head = root;
  
        // Recursively convert left subtree
        BToDLL(root.left);
    }
  
    // Utility function for printing double linked list.
    void printList(Node head)
    {
        System.out.println("Extracted Double Linked List is : ");
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.right;
        }
    }
  
    // Driver program to test the above functions
    public static void main(String[] args)
    {
        /* Constructing below tree
               5
             /   \
            3     6
           / \     \
          1   4     8
         / \       / \
        0   2     7   9  */
  
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(5);
        tree.root.left = new Node(3);
        tree.root.right = new Node(6);
        tree.root.left.right = new Node(4);
        tree.root.left.left = new Node(1);
        tree.root.right.right = new Node(8);
        tree.root.left.left.right = new Node(2);
        tree.root.left.left.left = new Node(0);
        tree.root.right.right.left = new Node(7);
        tree.root.right.right.right = new Node(9);
  
        tree.BToDLL(tree.root);
        tree.printList(tree.head);
    }
}
  
// This code has been contributed by Mayank Jaiswal(mayank_24)
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# Python3 program to convert a given Binary Tree to Doubly Linked List 
class Node:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
  
class BinaryTree:
    # A simple recursive function to convert a given 
    # Binary tree to Doubly Linked List 
    # root    --> Root of Binary Tree 
    # head --> Pointer to head node of created doubly linked list 
    root, head = None, None
      
    def BToDll(self, root: Node):
        if root is None:
            return
  
        # Recursively convert right subtree
        self.BToDll(root.right)
  
        # Insert root into doubly linked list
        root.right = self.head
  
        # Change left pointer of previous head
        if self.head is not None:
            self.head.left = root
  
        # Change head of doubly linked list
        self.head = root
  
        # Recursively convert left subtree
        self.BToDll(root.left)
  
    @staticmethod
    def print_list(head: Node):
        print('Extracted Double Linked list is:')
        while head is not None:
            print(head.data, end = ' ')
            head = head.right
  
# Driver program to test above function 
if __name__ == '__main__':
      
    """
    Constructing below tree
            5
        // \\
        3 6
        // \\ \\
        1 4 8
    // \\ // \\
    0 2 7 9
    """
    tree = BinaryTree()
    tree.root = Node(5)
    tree.root.left = Node(3)
    tree.root.right = Node(6)
    tree.root.left.left = Node(1)
    tree.root.left.right = Node(4)
    tree.root.right.right = Node(8)
    tree.root.left.left.left = Node(0)
    tree.root.left.left.right = Node(2)
    tree.root.right.right.left = Node(7)
    tree.root.right.right.right = Node(9)
  
    tree.BToDll(tree.root)
    tree.print_list(tree.head)
  
# This code is contributed by Rajat Srivastava
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// C# program to convert a given Binary Tree to Doubly Linked List
using System;
  
/* Structure for tree and Linked List */
public class Node {
    public int data;
    public Node left, right;
  
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
  
class GFG {
    // root    --> Root of Binary Tree
    public Node root;
  
    // head --> Pointer to head node of created doubly linked list
    public Node head;
  
    // A simple recursive function to convert a given
    // Binary tree to Doubly Linked List
    public virtual void BToDLL(Node root)
    {
        // Base cases
        if (root == null)
            return;
  
        // Recursively convert right subtree
        BToDLL(root.right);
  
        // insert root into DLL
        root.right = head;
  
        // Change left pointer of previous head
        if (head != null)
            head.left = root;
  
        // Change head of Doubly linked list
        head = root;
  
        // Recursively convert left subtree
        BToDLL(root.left);
    }
  
    // Utility function for printing double linked list.
    public virtual void printList(Node head)
    {
        Console.WriteLine("Extracted Double "
                          + "Linked List is : ");
        while (head != null) {
            Console.Write(head.data + " ");
            head = head.right;
        }
    }
  
    // Driver program to test above function
    public static void Main(string[] args)
    {
        /* Constructing below tree 
            
            / \ 
            3     6 
        / \     \ 
        1 4     8 
        / \     / \ 
        0 2     7 9 */
  
        GFG tree = new GFG();
        tree.root = new Node(5);
        tree.root.left = new Node(3);
        tree.root.right = new Node(6);
        tree.root.left.right = new Node(4);
        tree.root.left.left = new Node(1);
        tree.root.right.right = new Node(8);
        tree.root.left.left.right = new Node(2);
        tree.root.left.left.left = new Node(0);
        tree.root.right.right.left = new Node(7);
        tree.root.right.right.right = new Node(9);
  
        tree.BToDLL(tree.root);
        tree.printList(tree.head);
    }
}
  
// This code is contributed by Shrikant13
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Output :



Extracted Double Linked list is:
0 1 2 3 4 5 6 7 8 9

Time Complexity: O(n), as the solution does a single traversal of given Binary Tree.

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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