Convert Binary Tree to Doubly Linked List by keeping track of visited node (original) (raw)

Last Updated : 23 Jul, 2025

Given a Binary Tree, The task is to convert it to a **Doubly Linked List keeping the same order.

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 I**norder 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 a Doubly Linked List | Set 1
Convert a given Binary Tree to a Doubly Linked List | Set 2

**Approach: Below is the idea to solve the problem:

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 set previous of this node as **prev.

Below is the implementation of the above approach:

C++ `

// A C++ program for in-place conversion of Binary Tree to // DLL #include 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 Code 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;

// 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

// 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) { console.log(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

**Note: The use of static variables like above is not a recommended practice, here static is used for simplicity. Imagine if 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 a reference to a pointer.

**Time Complexity: O(N), The above program does a simple inorder traversal, so time complexity is O(N) where N is the number of nodes in a given Binary tree.
**Auxiliary Space: O(N), For recursion call stack.

Convert a given Binary Tree to Doubly Linked List iteratively using Stack data structure:

Do iterative inorder traversal and maintain a **prev pointer to point the last visited node then point **current node's perv to prev and **prev's next to current node.

Below is the implementation of the above approach:

C++ `

// A C++ program for in-place conversion of Binary Tree to // DLL #include <bits/stdc++.h> using namespace std;

/* A binary tree node has data, and left and right pointers / struct node { int data; node left; node* right; };

node * bToDLL(node root) { stack<pair<node*, int>> s; s.push({root, 0}); vector 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; }

/* Function to print nodes in a given doubly linked list / void printList(node node) { while (node != NULL) { cout << node->data << " "; node = node->right; } }

// Convert to DLL
node* head = bToDLL(root);

// Print the converted list
printList(head);

return 0;

}

Java

import java.util.Stack;

class Node { int data; Node left; Node right;

/* Helper function that allocates a new node with the given data and NULL left and right pointers. */ public Node(int data) { this.data = data; left = null; right = null; } }

public class Main { static Node bToDLL(Node root) { Stack<Pair<Node, Integer>> s = new Stack<>(); s.push(new Pair<>(root, 0)); boolean flag = true; Node head = null; Node prev = null; while (!s.empty()) { Pair<Node, Integer> x = s.pop(); Node t = x.getKey(); int state = x.getValue(); 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; }

/* Function to print nodes in a given doubly linked list */ static void printList(Node head) { while (head != null) { System.out.print(head.data + " "); head = head.right; } }

public static void main(String[] args) 
{
  
    // Let us create the tree shown in above diagram
    Node 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
    Node head = bToDLL(root);

    // Print the converted list
    printList(head);
}

}

class Pair<K, V> { private K key; private V value;

public Pair(K key, V value) {
    this.key = key;
    this.value = value;
}

public K getKey() {
    return key;
}

public V getValue() {
    return value;
}

}

// This code is contributed by aadityamaharshi21.

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#

// C# code implementation for the above approach

using System; using System.Collections.Generic;

public class Node { public int data; public Node left; public Node right;

/* Helper function that allocates a new node with the given data and NULL left and right pointers. */ public Node(int data) { this.data = data; left = null; right = null; } }

public class GFG {

static Node bToDLL(Node root) { Stack<KeyValuePair<Node, int> > s = new Stack<KeyValuePair<Node, int> >(); s.Push(new KeyValuePair<Node, int>(root, 0)); bool flag = true; Node head = null; Node prev = null; while (s.Count != 0) { KeyValuePair<Node, int> x = s.Pop(); Node t = x.Key; int state = x.Value; if (state == 3 || t == null) { continue; } s.Push( new KeyValuePair<Node, int>(t, state + 1)); if (state == 0) { s.Push( new KeyValuePair<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 KeyValuePair<Node, int>(t.right, 0)); } } return head; }

/* Function to print nodes in a given doubly linked list */ static void printList(Node head) { while (head != null) { System.Console.Write(head.data + " "); head = head.right; } }

static public void Main() {

// Code
// Let us create the tree shown in above diagram
Node 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
Node head = bToDLL(root);

// Print the converted list
printList(head);

} }

// This code is contributed by lokesh.

JavaScript

// JavaScript code for above approach

class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } }

function bToDLL(root) { let s = []; s.push([root, 0]); let res = []; let flag = true; let head = null; let prev = null;

while (s.length > 0) {
  let [t, state] = s.pop();
  if (state === 3 || 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;

}

function newNode(data) { let new_node = new Node(data); return new_node; }

function printList(node) { while (node !== null) { console.log(node.data); node = node.right; } }

let 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);

let head = bToDLL(root); printList(head);

// This code is contributed by adityamaharshi21

**Time complexity: O(N)
**Auxiliary Space: O(N)

Convert a given Binary Tree to Doubly Linked List iteratively using Morris Traversal:

The idea is to keep track of previous node while doing Inorder tree traversal using Morris Traversal. This removes the need for a recursion call stack or a stack thus **reducing space complexity

C++ `

#include

// Structure for a binary tree node struct Node { int data; Node* left; Node* right;

// Constructor
Node(int val) : data(val), left(nullptr), right(nullptr) {}

};

// Function to convert binary tree to doubly linked list Node* bToDLL(Node* root) { // Initialization Node* curr = root; Node* prev = nullptr; // Used to keep track of prev node Node* final_head = nullptr; // Used to return the final head

// Traverse the tree
while (curr) {
    // If no left child
    if (!curr->left) {
        // If final_head is not set, set it to current node
        if (!final_head) {
            prev = curr;
            final_head = curr;
        } else {
            // Set next of prev as curr and prev of curr as prev
            prev->right = curr;
            curr->left = prev;
        }
        // Set the new prev node
        prev = curr;
        curr = curr->right;
    } else {
        // If left child exists
        Node* pre = curr->left;
        while (pre->right && pre->right != curr) {
            pre = pre->right;
        }
        if (!pre->right) {
            pre->right = curr;
            curr = curr->left;
        } else {
            curr = pre->right;
            // Set next of prev as curr and prev of curr as prev
            prev->right = curr;
            curr->left = prev;
            // Set the new prev node
            prev = curr;
            curr = curr->right;
        }
    }
}

return final_head; // Return the final head of the linked list

}

// Function to print nodes in the doubly linked list void print_dll(Node* head) { while (head != nullptr) { std::cout << head->data << " "; head = head->right; } }

// Driver code int main() { // Create the binary tree Node* 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 the binary tree to doubly linked list
Node* head = bToDLL(root);

// Print the converted list
print_dll(head);

return 0;

}

Java

// Definition of Node class representing a tree node class Node { int data; Node left, right;

public Node(int item) {
    data = item;
    left = right = null;
}

}

// Class to convert a binary tree to a doubly linked list public class BinaryTreeToDLL { // Function to convert a binary tree to a doubly linked list static Node bToDLL(Node root) { Node curr = root; Node prev = null; // Used to keep track of prev node Node finalHead = null; // Used to return the final head

    // Traverse the binary tree
    while (curr != null) {
        // If current node has no left child
        if (curr.left == null) {
            // If finalHead is not set, set it to current node
            if (finalHead == null) {
                prev = curr;
                finalHead = curr;
            } else {
                // Connect previous node's right to current node
                prev.right = curr;
                // Connect current node's left to previous node
                curr.left = prev;
            }
            // Move prev pointer to current node
            prev = curr;
            // Move current pointer to its right child
            curr = curr.right;
        } else {
            // If current node has left child, find its inorder predecessor
            Node pre = curr.left;
            while (pre.right != null && pre.right != curr) {
                pre = pre.right;
            }
            if (pre.right == null) {
                // If inorder predecessor's right is not connected, connect it to current node
                pre.right = curr;
                // Move current pointer to its left child
                curr = curr.left;
            } else {
                // If inorder predecessor's right is connected,
                // connect current node to previous node and move pointers
                curr = pre.right;
                prev.right = curr;
                curr.left = prev;
                prev = curr;
                curr = curr.right;
            }
        }
    }

    // Return the final head of the doubly linked list
    return finalHead;
}

// Function to print the doubly linked list
static void printDLL(Node head) {
    StringBuilder result = new StringBuilder();
    // Concatenate nodes data into a single line 
    while (head != null) {
        result.append(head.data).append(" ");
        head = head.right;
    }
    System.out.println("Elements are: " + result.toString().trim());
}

// Driver code to test above functions 
public static void main(String args[]) {
    // Create the tree as shown in the diagram 
    Node 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 the binary tree to doubly linked list
    Node head = bToDLL(root);

    // Print the converted list 
    printDLL(head);
}

}

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

def bToDLL(root): # do Code here curr = root prev = None #Used to keep track of prev node final_head = None #used to return the final head while curr: if not curr.left: if not final_head: prev = curr final_head = curr else: #set next of prev as curr and prev of curr as prev prev.right = curr curr.left = prev #set the new prev node prev = curr curr = curr.right else: pre = curr.left while pre.right and pre.right != curr: pre = pre.right if not pre.right: pre.right = curr curr = curr.left else: curr = pre.right #set next of prev as curr and prev of curr as prev prev.right = curr curr.left = prev #set the new prev node prev = curr curr = curr.right

    return final_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 = bToDLL(root) 
  
# Print the converted list 
print_dll(head)

JavaScript

class Node { constructor(val) { this.data = val; this.left = null; this.right = null; } }

function bToDLL(root) { let curr = root; let prev = null; // Used to keep track of prev node let final_head = null; // Used to return the final head

while (curr) {
    if (!curr.left) {
        if (!final_head) {
            prev = curr;
            final_head = curr;
        } else {
            prev.right = curr;
            curr.left = prev;
        }
        prev = curr;
        curr = curr.right;
    } else {
        let pre = curr.left;
        while (pre.right && pre.right !== curr) {
            pre = pre.right;
        }
        if (!pre.right) {
            pre.right = curr;
            curr = curr.left;
        } else {
            curr = pre.right;
            prev.right = curr;
            curr.left = prev;
            prev = curr;
            curr = curr.right;
        }
    }
}

return final_head;

}

function printDLL(head) { let result = ""; // Function to concatenate nodes data into a single line while (head !== null) { result += head.data + " "; head = head.right; } console.log(result.trim()); }

// Driver program to test above functions // Let us create the tree as shown in above diagram

let 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);

let head = bToDLL(root);

// Print the converted list printDLL(head);

**Time complexity: O(N)

**Auxiliary Space: O(1)