Convert Binary Tree to Doubly Linked List using inorder traversal (original) (raw)
Last Updated : 23 Jul, 2025
Given a **Binary Tree (BT), the task is to convert it to a **Doubly Linked List (DLL) in place. The **left and **right pointers in nodes will be used as **previous and **next pointers respectively in converted DLL. The order of nodes in **DLL must be the same as the order of the given Binary Tree. The first node of **Inorder traversal (leftmost node in BT) must be the head node of the DLL.
**Examples:
**Input:
**Output:
**Explanation: _The above binary tree is converted into doubly linked list where left pointer of the binary tree node act as the previous node and right pointer of the binary tree node act as the next node.
**Input:
**Output:
**Explanation: _The above binary tree is converted into doubly linked list where left pointer of the binary tree node act as the previous node and right pointer of the binary tree node act as the next node.
Table of Content
- [Naive Approach] Using Recursion - O(n) Time and O(h) Space
- [Expected Approach] Using Morris Traversal Algorithm - O(n) Time and O(1) Space
[Naive Approach] Using Recursion - O(n) Time and O(h) Space
The idea is to recursively traverse the binary tree using **inorder traversal. At each node, if left subtree exists, find the**inorder predecessor, then process the left subtree and link the **current node and **predecessor. If right subtree exists, then find the**inorder successor, process the right subtree and then link the current node and **successor node.
Below is the implementation of the above approach:
C++ `
// C++ program for in-place // conversion of Binary Tree to DLL #include <bits/stdc++.h> using namespace std;
class Node { public: int data; Node* left; Node* right; Node (int x) { data = x; left = nullptr; right = nullptr; } };
void inorder(Node* root) {
// if left subtree exists
if (root->left){
// find the inorder predecessor of root node
Node* pred = root->left;
while (pred->right){
pred = pred->right;
}
// process the left subtree
inorder(root->left);
// link the predecessor and root node
pred->right = root;
root->left = pred;
}
// if right subtree exists
if (root->right) {
// find the inorder successor of root node
Node* succ = root->right;
while (succ->left) {
succ = succ->left;
}
// process the right subtree
inorder(root->right);
// link the successor and root node
root->right = succ;
succ->left = root;
}}
Node* bToDLL(Node* root){
// return if root is null
if (root == nullptr) return root;
// find the head of dll
Node* head = root;
while (head->left != nullptr)
head = head->left;
// recursively convert the tree into dll
inorder(root);
return head;}
void printList(Node* head){ Node* curr = head;
while (curr != NULL) {
cout << curr->data << " ";
curr = curr->right;
}
cout<<endl;}
int main() {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node* head = bToDLL(root);
printList(head);
return 0;}
C
// C program for in-place // conversion of Binary Tree to DLL #include <stdio.h> #include <stdlib.h>
struct Node { int data; struct Node* left; struct Node* right; };
// Inorder traversal to link nodes void inorder(struct Node* root) {
// if left subtree exists
if (root->left) {
// find the inorder predecessor of root node
struct Node* pred = root->left;
while (pred->right) {
pred = pred->right;
}
// process the left subtree
inorder(root->left);
// link the predecessor and root node
pred->right = root;
root->left = pred;
}
// if right subtree exists
if (root->right) {
// find the inorder successor of root node
struct Node* succ = root->right;
while (succ->left) {
succ = succ->left;
}
// process the right subtree
inorder(root->right);
// link the successor and root node
root->right = succ;
succ->left = root;
}}
// Function to convert binary tree to doubly linked list struct Node* bToDLL(struct Node* root) {
// return if root is null
if (root == NULL) return root;
// find the head of dll
struct Node* head = root;
while (head->left != NULL)
head = head->left;
// recursively convert the tree into dll
inorder(root);
return head;}
void printList(struct Node* head) { struct Node* curr = head;
while (curr != NULL) {
printf("%d ", curr->data);
curr = curr->right;
}
printf("\n");}
struct Node* createNode(int x) { struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->data = x; node->left = NULL; node->right = NULL; return node; }
int main() {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
struct Node* root = createNode(10);
root->left = createNode(12);
root->right = createNode(15);
root->left->left = createNode(25);
root->left->right = createNode(30);
root->right->left = createNode(36);
struct Node* head = bToDLL(root);
printList(head);
return 0;}
Java
// Java program for in-place // conversion of Binary Tree to DLL class Node { int data; Node left, right;
Node(int x) {
data = x;
left = null;
right = null;
}}
class GfG {
// Inorder traversal to link nodes
static void inorder(Node root) {
// if left subtree exists
if (root.left != null) {
// find the inorder predecessor of root node
Node pred = root.left;
while (pred.right != null) {
pred = pred.right;
}
// process the left subtree
inorder(root.left);
// link the predecessor and root node
pred.right = root;
root.left = pred;
}
// if right subtree exists
if (root.right != null) {
// find the inorder successor of root node
Node succ = root.right;
while (succ.left != null) {
succ = succ.left;
}
// process the right subtree
inorder(root.right);
// link the successor and root node
root.right = succ;
succ.left = root;
}
}
// Function to convert binary tree to doubly linked list
static Node bToDLL(Node root) {
// return if root is null
if (root == null) return root;
// find the head of dll
Node head = root;
while (head.left != null)
head = head.left;
// recursively convert the tree into dll
inorder(root);
return head;
}
static void printList(Node head) {
Node curr = head;
while (curr != null) {
System.out.print(curr.data + " ");
curr = curr.right;
}
System.out.println();
}
public static void main(String[] args) {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node head = bToDLL(root);
printList(head);
}}
Python
Python program for in-place
conversion of Binary Tree to DLL
class Node: def init(self, new_value): self.data = new_value self.left = None self.right = None
Inorder traversal to link nodes
def inorder(root):
# if left subtree exists
if root.left:
# find the inorder predecessor of root node
pred = root.left
while pred.right:
pred = pred.right
# process the left subtree
inorder(root.left)
# link the predecessor and root node
pred.right = root
root.left = pred
# if right subtree exists
if root.right:
# find the inorder successor of root node
succ = root.right
while succ.left:
succ = succ.left
# process the right subtree
inorder(root.right)
# link the successor and root node
root.right = succ
succ.left = rootFunction to convert binary tree to doubly linked list
def bToDLL(root):
# return if root is null
if root is None:
return root
# find the head of dll
head = root
while head.left:
head = head.left
# recursively convert the tree into dll
inorder(root)
return headdef print_list(head): curr = head while curr: print(curr.data, end=" ") curr = curr.right print()
if name == "main":
# Create a hard coded binary tree
# 10
# / \
# 12 15
# / \ /
# 25 30 36
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_list(head)C#
// C# program for in-place // conversion of Binary Tree to DLL using System;
class Node { public int data; public Node left, right;
public Node(int x) {
data = x;
left = right = null;
}}
class GfG {
// Inorder traversal to link nodes
static void Inorder(Node root) {
// if left subtree exists
if (root.left != null) {
// find the inorder predecessor of root node
Node pred = root.left;
while (pred.right != null) {
pred = pred.right;
}
// process the left subtree
Inorder(root.left);
// link the predecessor and root node
pred.right = root;
root.left = pred;
}
// if right subtree exists
if (root.right != null) {
// find the inorder successor of root node
Node succ = root.right;
while (succ.left != null) {
succ = succ.left;
}
// process the right subtree
Inorder(root.right);
// link the successor and root node
root.right = succ;
succ.left = root;
}
}
// Function to convert binary tree to doubly linked list
static Node BToDLL(Node root) {
// return if root is null
if (root == null) return root;
// find the head of dll
Node head = root;
while (head.left != null)
head = head.left;
// recursively convert the tree into dll
Inorder(root);
return head;
}
static void PrintList(Node head) {
Node curr = head;
while (curr != null) {
Console.Write(curr.data + " ");
curr = curr.right;
}
Console.WriteLine();
}
static void Main(string[] args) {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node head = BToDLL(root);
PrintList(head);
}}
JavaScript
// JavaScript program for in-place // conversion of Binary Tree to DLL class Node { constructor(new_value) { this.data = new_value; this.left = null; this.right = null; } }
// Inorder traversal to link nodes function inorder(root) {
// if left subtree exists
if (root.left) {
// find the inorder predecessor of root node
let pred = root.left;
while (pred.right) {
pred = pred.right;
}
// process the left subtree
inorder(root.left);
// link the predecessor and root node
pred.right = root;
root.left = pred;
}
// if right subtree exists
if (root.right) {
// find the inorder successor of root node
let succ = root.right;
while (succ.left) {
succ = succ.left;
}
// process the right subtree
inorder(root.right);
// link the successor and root node
root.right = succ;
succ.left = root;
}}
// Function to convert binary tree to doubly linked list function bToDLL(root) {
// return if root is null
if (root === null) return root;
// find the head of dll
let head = root;
while (head.left !== null)
head = head.left;
// recursively convert the tree into dll
inorder(root);
return head;}
function printList(head) { let curr = head;
while (curr !== null) {
console.log(curr.data);
curr = curr.right;
}}
// Create a hard coded binary tree
// 10
// /
// 12 15
// / \ /
// 25 30 36
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);
printList(head);
`
**Time Complexity: O(n), where **n is the number of node in the tree.
**Auxiliary Space: O(h), where **h is the height of tree.
[Expected Approach] Using Morris Traversal Algorithm - O(n) Time and O(1) Space
The idea is to use **Morris traversal algorithmto traverse the binary tree, while maintaining proper linkages between the nodes.
Step by step implementation:
- Initialize pointers **head and **tail. head will point to the head node of the resultant **dll and tail will point to the last node in **dll.
- Initialize another pointer **curr, which will initially point to **root node. Start traversing until curr is not NULL
- If curr.left is **null, then add the current node to the list (If head is **empty, then make this node as **head node) and move **curr to curr.right.
- If **curr.left is not **null, then find the **inorder predecessor of the **current node. Let that node be ****'pred'**. There are two possibilites:
- If pred.right is equal to **null, then create a link between **pred and **curr, by setting **pred.right = curr and set **curr = curr.left.
- If **pred->right is equal to **curr, then this means we have traversed the **left subtree and now we can add the curr node to the list. Then set curr = curr->right.
- Return the **head. C++ `
// C++ program for in-place // conversion of Binary Tree to DLL #include <bits/stdc++.h> using namespace std;
class Node { public: int data; Node *left; Node *right; Node(int x) { data = x; left = nullptr; right = nullptr; } };
// Function to perform Morris Traversal and convert // binary tree to doubly linked list (DLL) Node* morrisTraversal(Node* root) {
// return if root is null
if (root == nullptr) return root;
// head and tail node for the dll
Node* head = nullptr, *tail = nullptr;
Node* curr = root;
while (curr != nullptr) {
// if left tree does not exists,
// then add the curr node to the
// dll and set curr = curr->right
if (curr->left == nullptr) {
if (head == nullptr) {
head = tail = curr;
}
else {
tail->right = curr;
curr->left = tail;
tail = curr;
}
curr = curr->right;
}
else {
Node* pred = curr->left;
// find the inorder predecessor
while (pred->right != nullptr && pred->right != curr) {
pred = pred->right;
}
// create a linkage between pred and
// curr
if (pred->right == nullptr) {
pred->right = curr;
curr = curr->left;
}
// if pred->right = curr, it means
// we have processed the left subtree,
// and we can add curr node to list
else {
tail->right = curr;
curr->left = tail;
tail = curr;
curr = curr->right;
}
}
}
return head;}
void printList(Node* head) { Node* curr = head;
while (curr != nullptr) {
cout << curr->data << " ";
curr = curr->right;
}
cout << endl;}
int main() {
// Create a hard-coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node* head = morrisTraversal(root);
printList(head);
return 0;}
C
// C program for in-place // conversion of Binary Tree to DLL #include <stdio.h> #include <stdlib.h>
struct Node { int data; struct Node* left; struct Node* right; };
struct Node* morrisTraversal(struct Node* root) {
// return if root is null
if (root == NULL) return root;
// head and tail node for the dll
struct Node* head = NULL, *tail = NULL;
struct Node* curr = root;
while (curr != NULL) {
// if left tree does not exists,
// then add the curr node to the
// dll and set curr = curr->right
if (curr->left == NULL) {
if (head == NULL) {
head = tail = curr;
}
else {
tail->right = curr;
curr->left = tail;
tail = curr;
}
curr = curr->right;
}
else {
struct Node* pred = curr->left;
// find the inorder predecessor
while (pred->right != NULL
&& pred->right != curr) {
pred = pred->right;
}
// create a linkage between pred and
// curr
if (pred->right == NULL) {
pred->right = curr;
curr = curr->left;
}
// if pred->right = curr, it means
// we have processed the left subtree,
// and we can add curr node to list
else {
tail->right = curr;
curr->left = tail;
tail = curr;
curr = curr->right;
}
}
}
return head;}
void printList(struct Node* head) { struct Node* curr = head;
while (curr != NULL) {
printf("%d ", curr->data);
curr = curr->right;
}
printf("\n");}
struct Node* createNode(int new_value) { struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->data = new_value; node->left = node->right = NULL; return node; }
int main() {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
struct Node* root = createNode(10);
root->left = createNode(12);
root->right = createNode(15);
root->left->left = createNode(25);
root->left->right = createNode(30);
root->right->left = createNode(36);
struct Node* head = morrisTraversal(root);
printList(head);
return 0;}
Java
// Java program for in-place // conversion of Binary Tree to DLL
class Node { int data; Node left, right;
Node(int x) {
data = x;
left = null;
right = null;
}}
class GfG {
static Node morrisTraversal(Node root) {
// return if root is null
if (root == null) return root;
// head and tail node for the dll
Node head = null, tail = null;
Node curr = root;
while (curr != null) {
// if left tree does not exists,
// then add the curr node to the
// dll and set curr = curr.right
if (curr.left == null) {
if (head == null) {
head = tail = curr;
}
else {
tail.right = curr;
curr.left = tail;
tail = curr;
}
curr = curr.right;
} else {
Node pred = curr.left;
// find the inorder predecessor
while (pred.right != null
&& pred.right != curr) {
pred = pred.right;
}
// create a linkage between pred and
// curr
if (pred.right == null) {
pred.right = curr;
curr = curr.left;
}
// if pred.right = curr, it means
// we have processed the left subtree,
// and we can add curr node to list
else {
tail.right = curr;
curr.left = tail;
tail = curr;
curr = curr.right;
}
}
}
return head;
}
static void printList(Node head) {
Node curr = head;
while (curr != null) {
System.out.print(curr.data + " ");
curr = curr.right;
}
System.out.println();
}
public static void main(String[] args) {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node head = morrisTraversal(root);
printList(head);
}}
Python
Python program for in-place
conversion of Binary Tree to DLL
class Node: def init(self, new_value): self.data = new_value self.left = None self.right = None
def morris_traversal(root):
# return if root is None
if root is None:
return root
# head and tail node for the dll
head = None
tail = None
curr = root
while curr is not None:
# if left tree does not exist,
# then add the curr node to the
# dll and set curr = curr.right
if curr.left is None:
if head is None:
head = tail = curr
else:
tail.right = curr
curr.left = tail
tail = curr
curr = curr.right
else:
pred = curr.left
# find the inorder predecessor
while pred.right is not None \
and pred.right != curr:
pred = pred.right
# create a linkage between pred and curr
if pred.right is None:
pred.right = curr
curr = curr.left
# if pred.right = curr, it means
# we have processed the left subtree,
# and we can add curr node to list
else:
tail.right = curr
curr.left = tail
tail = curr
curr = curr.right
return headdef print_list(head): curr = head
while curr is not None:
print(curr.data, end=" ")
curr = curr.right
print()if name == "main":
# Create a hard coded binary tree
# 10
# / \
# 12 15
# / \ /
# 25 30 36
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 = morris_traversal(root)
print_list(head)C#
// C# program for in-place // conversion of Binary Tree to DLL
using System;
class Node { public int data; public Node left, right;
public Node(int x) {
data = x;
left = null;
right = null;
}}
class GfG {
static Node MorrisTraversal(Node root) {
// return if root is null
if (root == null) return root;
// head and tail node for the dll
Node head = null, tail = null;
Node curr = root;
while (curr != null) {
// if left tree does not exists,
// then add the curr node to the
// dll and set curr = curr.right
if (curr.left == null) {
if (head == null) {
head = tail = curr;
} else {
tail.right = curr;
curr.left = tail;
tail = curr;
}
curr = curr.right;
} else {
Node pred = curr.left;
// find the inorder predecessor
while (pred.right != null
&& pred.right != curr) {
pred = pred.right;
}
// create a linkage between pred and
// curr
if (pred.right == null) {
pred.right = curr;
curr = curr.left;
}
// if pred.right = curr, it means
// we have processed the left subtree,
// and we can add curr node to list
else {
tail.right = curr;
curr.left = tail;
tail = curr;
curr = curr.right;
}
}
}
return head;
}
static void PrintList(Node head) {
Node curr = head;
while (curr != null) {
Console.Write(curr.data + " ");
curr = curr.right;
}
Console.WriteLine();
}
static void Main(string[] args) {
// Create a hard coded binary tree
// 10
// / \
// 12 15
// / \ /
// 25 30 36
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);
Node head = MorrisTraversal(root);
PrintList(head);
}}
JavaScript
// JavaScript program for in-place // conversion of Binary Tree to DLL
class Node { constructor(new_value) { this.data = new_value; this.left = this.right = null; } }
function morrisTraversal(root) {
// return if root is null
if (root === null) return root;
// head and tail node for the dll
let head = null, tail = null;
let curr = root;
while (curr !== null) {
// if left tree does not exists,
// then add the curr node to the
// dll and set curr = curr.right
if (curr.left === null) {
if (head === null) {
head = tail = curr;
} else {
tail.right = curr;
curr.left = tail;
tail = curr;
}
curr = curr.right;
} else {
let pred = curr.left;
// find the inorder predecessor
while (pred.right !== null && pred.right !== curr) {
pred = pred.right;
}
// create a linkage between pred and curr
if (pred.right === null) {
pred.right = curr;
curr = curr.left;
}
// if pred.right = curr, it means
// we have processed the left subtree,
// and we can add curr node to list
else {
tail.right = curr;
curr.left = tail;
tail = curr;
curr = curr.right;
}
}
}
return head;}
function printList(head) { let curr = head;
while (curr !== null) {
console.log(curr.data);
curr = curr.right;
}}
// Create a hard coded binary tree
// 10
// /
// 12 15
// / \ /
// 25 30 36
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 = morrisTraversal(root);
printList(head);
`
**Time Complexity: O(n), where **n is the number of nodes in tree.
**Auxiliary Space: O(1)