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 Inorder for the given Binary Tree.
- The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL.
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 <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 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; // 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: 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<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; } /* 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 = 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 |
Output
25 12 30 10 36 15
Time complexity: O(N)
Auxiliary Space: O(N)
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