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Convert a given Binary Tree to Doubly Linked List | Set 3

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  • Difficulty Level : Medium
  • Last Updated : 01 Aug, 2022

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 the same as in Inorder for the given Binary Tree. The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL. 

TreeToList

The following two 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

In this post, a third solution is discussed which seems to be the simplest of all. The idea is to do in order traversal of the binary tree. While doing inorder traversal, keep track of the previously visited node in a variable, say prev. For every visited node, make it next to the prev and previous of this node as prev.

 

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The following is the implementation of this solution. 

C++




// A C++ program for in-place conversion of Binary Tree to
// DLL
#include <iostream>
using namespace std;
 
/* A binary tree node has data, and left and right pointers
 */
struct node {
    int data;
    node* left;
    node* right;
};
 
// 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 BinaryTree2DoubleLinkedList(node* root, node** head)
{
    // Base case
    if (root == NULL)
        return;
 
    // Initialize previously visited node as NULL. This is
    // static so that the same value is accessible in all
    // recursive calls
    static node* prev = NULL;
 
    // Recursively convert left subtree
    BinaryTree2DoubleLinkedList(root->left, head);
 
    // Now convert this node
    if (prev == NULL)
        *head = root;
    else {
        root->left = prev;
        prev->right = root;
    }
    prev = root;
 
    // Finally convert right subtree
    BinaryTree2DoubleLinkedList(root->right, head);
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
node* newNode(int data)
{
    node* new_node = new node;
    new_node->data = data;
    new_node->left = new_node->right = NULL;
    return (new_node);
}
 
/* Function to print nodes in a given doubly linked list */
void printList(node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->right;
    }
}
 
/* Driver program to test above functions*/
int main()
{
    // Let us create the tree shown in above diagram
    node* root = newNode(10);
    root->left = newNode(12);
    root->right = newNode(15);
    root->left->left = newNode(25);
    root->left->right = newNode(30);
    root->right->left = newNode(36);
 
    // Convert to DLL
    node* head = NULL;
    BinaryTree2DoubleLinkedList(root, &head);
 
    // Print the converted list
    printList(head);
 
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

C




// A C program for in-place conversion of Binary Tree to DLL
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, and left and right pointers
 */
typedef struct node {
    int data;
    struct node* left;
    struct node* right;
} 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 BinaryTree2DoubleLinkedList(node* root, node** head)
{
    // Base case
    if (root == NULL)
        return;
 
    // Initialize previously visited node as NULL. This is
    // static so that the same value is accessible in all
    // recursive calls
    static node* prev = NULL;
 
    // Recursively convert left subtree
    BinaryTree2DoubleLinkedList(root->left, head);
 
    // Now convert this node
    if (prev == NULL)
        *head = root;
    else {
        root->left = prev;
        prev->right = root;
    }
    prev = root;
 
    // Finally convert right subtree
    BinaryTree2DoubleLinkedList(root->right, head);
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
node* newNode(int data)
{
    node* new_node = (node*)malloc(sizeof(node));
    new_node->data = data;
    new_node->left = new_node->right = NULL;
    return (new_node);
}
 
/* Function to print nodes in a given doubly linked list */
void printList(node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->right;
    }
}
 
/* Driver program to test above functions*/
int main()
{
    // Let us create the tree shown in above diagram
    node* root = newNode(10);
    root->left = newNode(12);
    root->right = newNode(15);
    root->left->left = newNode(25);
    root->left->right = newNode(30);
    root->right->left = newNode(36);
 
    // Convert to DLL
    node* head = NULL;
    BinaryTree2DoubleLinkedList(root, &head);
 
    // Print the converted list
    printList(head);
 
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

Java




// A Java program for in-place conversion of Binary Tree to DLL
  
// A binary tree node has data, left pointers and right pointers
class Node
{
    int data;
    Node left, right;
  
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
  
class BinaryTree
{
    Node root;
      
    // head --> Pointer to head node of created doubly linked list
    Node head;
      
    // Initialize previously visited node as NULL. This is
    // static so that the same value is accessible in all recursive
    // calls
    static Node prev = null;
  
    // A simple recursive function to convert a given Binary tree
    // to Doubly Linked List
    // root --> Root of Binary Tree
    void BinaryTree2DoubleLinkedList(Node root)
    {
        // Base case
        if (root == null)
            return;
  
        // Recursively convert left subtree
        BinaryTree2DoubleLinkedList(root.left);
  
        // Now convert this node
        if (prev == null)
            head = root;
        else
        {
            root.left = prev;
            prev.right = root;
        }
        prev = root;
  
        // Finally convert right subtree
        BinaryTree2DoubleLinkedList(root.right);
    }
  
    /* Function to print nodes in a given doubly linked list */
    void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data + " ");
            node = node.right;
        }
    }
  
    // Driver program to test above functions
    public static void main(String[] args)
    {
        // Let us create the tree as shown in above diagram
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(10);
        tree.root.left = new Node(12);
        tree.root.right = new Node(15);
        tree.root.left.left = new Node(25);
        tree.root.left.right = new Node(30);
        tree.root.right.left = new Node(36);
  
        // convert to DLL
        tree.BinaryTree2DoubleLinkedList(tree.root);
          
        // Print the converted List
        tree.printList(tree.head);
  
    }
}
// This code has been contributed by Mayank Jaiswal(mayank_24)

Python3




# Python program for conversion of
# binary tree to doubly linked list.
class Node:
    def __init__(self, val):
        self.right = None
        self.data = val
        self.left = None
 
# Global variable used in convert
prev = None
 
def BinaryTree2DoubleLinkedList(root):
     
    # Base case
    if root is None:
        return root
             
    # Recursively convert left subtree
    head = BinaryTree2DoubleLinkedList(root.left);
     
    # Since we are going to change prev,
    # we need to use global keyword
    global prev
     
    # If prev is empty, then this is the
    # first node of DLL
    if prev is None :
        head = root
         
    else:
        root.left = prev
        prev.right = root
     
    # Update prev
    prev = root;
     
    # Recursively convert right subtree
    BinaryTree2DoubleLinkedList(root.right);
     
    return head
 
def print_dll(head):
     
    # Function to print nodes in given
    # doubly linked list
    while head is not None:
        print(head.data, end=" ")
        head = head.right
 
 
# Driver program to test above functions
# Let us create the tree as
# shown in above diagram
if __name__ == '__main__':
    root = Node(10)
    root.left = Node(12)
    root.right = Node(15)
    root.left.left = Node(25)
    root.left.right = Node(30)
    root.right.left = Node(36)
     
    head = BinaryTree2DoubleLinkedList(root)
     
    # Print the converted list
    print_dll(head)
     
# This code is contributed by codesankalp (SANKALP)

C#




// A C# program for in-place conversion
// of Binary Tree to DLL
using System;
 
// A binary tree node has data, left
// pointers and right pointers
public class Node
{
    public int data;
    public Node left, right;
 
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}
 
class GFG
{
public Node root;
 
// head --> Pointer to head node of
// created doubly linked list
public Node head;
 
// Initialize previously visited node
// as NULL. This is static so that the
// same value is accessible in all
// recursive calls
public static Node prev = null;
 
// A simple recursive function to
// convert a given Binary tree
// to Doubly Linked List
// root --> Root of Binary Tree
public virtual void BinaryTree2DoubleLinkedList(Node root)
{
    // Base case
    if (root == null)
    {
        return;
    }
 
    // Recursively convert left subtree
    BinaryTree2DoubleLinkedList(root.left);
 
    // Now convert this node
    if (prev == null)
    {
        head = root;
    }
    else
    {
        root.left = prev;
        prev.right = root;
    }
    prev = root;
 
    // Finally convert right subtree
    BinaryTree2DoubleLinkedList(root.right);
}
 
/* Function to print nodes in a
   given doubly linked list */
public virtual void printList(Node node)
{
    while (node != null)
    {
        Console.Write(node.data + " ");
        node = node.right;
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    // Let us create the tree as
    // shown in above diagram
    GFG tree = new GFG();
    tree.root = new Node(10);
    tree.root.left = new Node(12);
    tree.root.right = new Node(15);
    tree.root.left.left = new Node(25);
    tree.root.left.right = new Node(30);
    tree.root.right.left = new Node(36);
 
    // convert to DLL
    tree.BinaryTree2DoubleLinkedList(tree.root);
 
    // Print the converted List
    tree.printList(tree.head);
 
}
}
 
// This code is contributed by Shrikant13

Javascript




<script>
    // A javascript program for in-place conversion of Binary Tree to DLL
  
    // A binary tree node has data, left pointers and right pointers
    class Node {
        constructor(val) {
            this.data = val;
            this.left = null;
            this.right = null;
        }
    }
 
    var root;
      
    // head --> Pointer to head node of created doubly linked list
    var head;
      
    // Initialize previously visited node as NULL. This is
    //  so that the same value is accessible in all recursive
    // calls
    var prev = null;
  
    // A simple recursive function to convert a given Binary tree
    // to Doubly Linked List
    // root --> Root of Binary Tree
    function BinaryTree2DoubleLinkedList(root)
    {
        // Base case
        if (root == null)
            return;
  
        // Recursively convert left subtree
        BinaryTree2DoubleLinkedList(root.left);
  
        // Now convert this node
        if (prev == null)
            head = root;
        else
        {
            root.left = prev;
            prev.right = root;
        }
        prev = root;
  
        // Finally convert right subtree
        BinaryTree2DoubleLinkedList(root.right);
    }
  
    /* Function to print nodes in a given doubly linked list */
    function printList(node)
    {
        while (node != null)
        {
            document.write(node.data + " ");
            node = node.right;
        }
    }
  
    // Driver program to test above functions
 
    // Let us create the tree as shown in above diagram
      
    root = new Node(10);
    root.left = new Node(12);
    root.right = new Node(15);
    root.left.left = new Node(25);
    root.left.right = new Node(30);
    root.right.left = new Node(36);
  
    // convert to DLL
    BinaryTree2DoubleLinkedList(root);
         
    // Print the converted List
    printList(head);
  
// This code contributed by umadevi9616
</script>

Output

25 12 30 10 36 15 

Note that the use of static variables like above is not a recommended practice (we have used static for simplicity). Imagine a situation where the same function is called for two or more trees. The old value of prev would be used in the next call for a different tree. To avoid such problems, we can use a double-pointer or reference to a pointer.

Time Complexity: The above program does a simple inorder traversal, so time complexity is O(n) where n is the number of nodes in given binary tree.

The below code is the intuitive iterative implementation of the above inorder traversal approach (inspired from this video) –

C++




Node * bToDLL(Node *root)
{
    stack<pair<Node*, int>> s;
    s.push({root, 0});
    vector<int> res;
    bool flag = true;
    Node* head = NULL;
    Node* prev = NULL;
    while(!s.empty()) {
        auto x = s.top();
        Node* t = x.first;
        int state = x.second;
        s.pop();
        if(state == 3 or t == NULL) continue;
        s.push({t, state+1});
        if(state == 0) s.push({t->left, 0});
        else if(state == 1) {
            if(prev) prev->right = t;
            t->left = prev;
            prev = t;
            if(flag) {
                head = t;
                flag = false;
            }
        }
        else if(state == 2) s.push({t->right, 0});
    }
    return head;
}

Java




Node bToDLL(Node root)
{
    Stack<Pair> s = new Stack<Pair>();
    s.push(new Pair(root, 0));
    int[] res;
    boolean flag = true;
    Node head = null;
    Node prev = null;
    while (!s.empty()) {
        Pair x = s.peek();
        Node t = x.first;
        int state = x.second;
        s.pop();
        if (state == 3 || t == null)
            continue;
        s.push(new Pair(t, state + 1));
        if (state == 0)
            s.push(new Pair(t.left, 0));
        else if (state == 1) {
            if (prev != null)
                prev.right = t;
            t.left = prev;
            prev = t;
            if (flag) {
                head = t;
                flag = false;
            }
        }
        else if (state == 2)
            s.push(new Pair(t.right, 0));
    }
    return head;
}
 
// This code is contributed by Lovely Jain

Python3




def bToDLL(root):
    s = []
    s.append([root, 0])
    res = []
    flag = True
    head = None
    prev = None
    while len(s) > 0:
        x = s.pop()
        t = x[0]
        state = x[1]
        if state == 3 or t == None: continue
        s.append([t, state+1])
        if state == 0: s.append([t.left, 0])
        elif state == 1:
            if prev != None: prev.right = t
            t.left = prev
            prev = t
 
            if flag: 
                head = t
                flag = False 
 
        elif state == 2: s.append([t.right, 0])
 
    return head
   
  # This code is contributed by Tapeshdua420.

C#




Node bToDLL(Node root)
{
    Stack<Tuple<Node, int> > s
        = new Stack<Tuple<Node, int> >();
    s.Push(new Tuple<Node, int>(root, 0));
    List<int> res = new List<int>();
    bool flag = true;
    Node head = null;
    Node prev = null;
    while (s.Count > 0) {
        var x = s.Pop();
        Node t = x.Item1;
        int state = x.Item2;
        if (state == 3 || t == null)
            continue;
        s.Push(new Tuple<Node, int>(t, state + 1));
        if (state == 0)
            s.Push(new Tuple<Node, int>(t.left, 0));
        else if (state == 1) {
            if (prev != null)
                prev.right = t;
            t.left = prev;
            prev = t;
            if (flag) {
                head = t;
                flag = false;
            }
        }
        else if (state == 2)
            s.Push(new Tuple<Node, int>(t.right, 0));
    }
 
    return head;
}
 
// This code is contributed by Tapesh(tapeshdua420)

Time complexity: O(N)

Space complexity: O(N)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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