Given a Binary Tree, extract all leaves of it in a **D**oubly **L**inked **L**ist (DLL). Note that the DLL need to be created in-place. Assume that the node structure of DLL and Binary Tree is same, only the meaning of left and right pointers are different. In DLL, left means previous pointer and right means next pointer.

Let the following be input binary tree 1 / \ 2 3 / \ \ 4 5 6 / \ / \ 7 8 9 10 Output: Doubly Linked List 7<->8<->5<->9<->10 Modified Tree: 1 / \ 2 3 / \ 4 6

We need to traverse all leaves and connect them by changing their left and right pointers. We also need to remove them from Binary Tree by changing left or right pointers in parent nodes. There can be many ways to solve this. In the following implementation, we add leaves 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 then the left subtree. We use return values to update left or right pointers in parent nodes.

## C

// C program to extract leaves of a Binary Tree in a Doubly Linked List #include <stdio.h> #include <stdlib.h> // Structure for tree and linked list struct Node { int data; struct Node *left, *right; }; // Main function which extracts all leaves from given Binary Tree. // The function returns new root of Binary Tree (Note that root may change // if Binary Tree has only one node). The function also sets *head_ref as // head of doubly linked list. left pointer of tree is used as prev in DLL // and right pointer is used as next struct Node* extractLeafList(struct Node *root, struct Node **head_ref) { // Base cases if (root == NULL) return NULL; if (root->left == NULL && root->right == NULL) { // This node is going to be added to doubly linked list // of leaves, set right pointer of this node as previous // head of DLL. We don't need to set left pointer as left // is already NULL root->right = *head_ref; // Change left pointer of previous head if (*head_ref != NULL) (*head_ref)->left = root; // Change head of linked list *head_ref = root; return NULL; // Return new root } // Recur for right and left subtrees root->right = extractLeafList(root->right, head_ref); root->left = extractLeafList(root->left, head_ref); return root; } // Utility function for allocating node for Binary Tree. struct Node* newNode(int data) { struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->data = data; node->left = node->right = NULL; return node; } // Utility function for printing tree in In-Order. void print(struct Node *root) { if (root != NULL) { print(root->left); printf("%d ",root->data); print(root->right); } } // Utility function for printing double linked list. void printList(struct Node *head) { while (head) { printf("%d ", head->data); head = head->right; } } // Driver program to test above function int main() { struct Node *head = NULL; struct Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(6); root->left->left->left = newNode(7); root->left->left->right = newNode(8); root->right->right->left = newNode(9); root->right->right->right = newNode(10); printf("Inorder Trvaersal of given Tree is:\n"); print(root); root = extractLeafList(root, &head); printf("\nExtracted Double Linked list is:\n"); printList(head); printf("\nInorder traversal of modified tree is:\n"); print(root); return 0; }

## Java

// Java program to extract leaf nodes from binary tree // using double linked list // A binay tree node class Node { int data; Node left, right; Node(int item) { data = item; right = left = null; } } public class BinaryTree { Node root; Node head; // will point to head of DLL Node prev; // temporary pointer // The main fuction that links the list list to be traversed public Node extractLeafList(Node root) { if (root == null) return null; if (root.left == null && root.right == null) { if (head == null) { head = root; prev = root; } else { prev.right = root; root.left = prev; prev = root; } return null; } root.left = extractLeafList(root.left); root.right = extractLeafList(root.right); return root; } //Prints the DLL in both forward and reverse directions. public void printDLL(Node head) { Node last = null; while (head != null) { System.out.print(head.data + " "); last = head; head = head.right; } } void inorder(Node node) { if (node == null) return; inorder(node.left); System.out.print(node.data + " "); inorder(node.right); } // Driver program to test above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.right = new Node(6); tree.root.left.left.left = new Node(7); tree.root.left.left.right = new Node(8); tree.root.right.right.left = new Node(9); tree.root.right.right.right = new Node(10); System.out.println("Inorder traversal of given tree is : "); tree.inorder(tree.root); tree.extractLeafList(tree.root); System.out.println(""); System.out.println("Extracted double link list is : "); tree.printDLL(tree.head); System.out.println(""); System.out.println("Inorder traversal of modified tree is : "); tree.inorder(tree.root); } } // This code has been contributed by Mayank Jaiswal(mayank_24)

## Python

# Python program to extract leaf nodes from binary tree # using double linked list # A binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # Main function which extracts all leaves from given Binary Tree. # The function returns new root of Binary Tree (Note that # root may change if Binary Tree has only one node). # The function also sets *head_ref as head of doubly linked list. # left pointer of tree is used as prev in DLL # and right pointer is used as next def extractLeafList(root): # Base Case if root is None: return None if root.left is None and root.right is None: # This node is going to be added to doubly linked # list of leaves, set pointer of this node as # previous head of DLL. We don't need to set left # pointer as left is already None root.right = extractLeafList.head # Change the left pointer of previous head if extractLeafList.head is not None: extractLeafList.head.left = root # Change head of linked list extractLeafList.head = root return None # Return new root # Recur for right and left subtrees root.right = extractLeafList(root.right) root.left = extractLeafList(root.left) return root # Utility function for printing tree in InOrder def printInorder(root): if root is not None: printInorder(root.left) print root.data, printInorder(root.right) def printList(head): while(head): if head.data is not None: print head.data, head = head.right # Driver program to test above function extractLeafList.head = Node(None) root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.right = Node(6) root.left.left.left = Node(7) root.left.left.right = Node(8) root.right.right.left = Node(9) root.right.right.right = Node(10) print "Inorder traversal of given tree is:" printInorder(root) root = extractLeafList(root) print "\nExtract Double Linked List is:" printList(extractLeafList.head) print "\nInorder traversal of modified tree is:" printInorder(root)

Output:

Inorder Trvaersal of given Tree is: 7 4 8 2 5 1 3 9 6 10 Extracted Double Linked list is: 7 8 5 9 10 Inorder traversal of modified tree is: 4 2 1 3 6

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

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