Given a binary tree, the task is to convert it into a sorted linked list.
Examples:
Input:
1
/ \
2 3
Output: 1 2 3
Input:
2
/ \
4 8
/ \ / \
7 3 5 1
Output: 1 2 3 4 5 7 8
Input:
3
/
4
/
1
/
9
Output: 1 3 4 9
Approach: Recursively iterate the given binary tree and add each node to its correct position in the resultant linked list (initially empty) using insertion sort.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // A linked list node class Node { public: int data; Node* next; Node(int data) { this->data = data; this->next = NULL; } }; // A binary tree node class treeNode { public: int data; treeNode* left; treeNode* right; treeNode(int data) { this->data = data; this->left = NULL; this->right = NULL; } }; // Function to print the linked list void print(Node* head) { if (head == NULL) { return; } Node* temp = head; while (temp != NULL) { cout << temp->data << " "; temp = temp->next; } } // Function to create Linked list from given binary tree Node* sortedList(Node* head, treeNode* root) { // return head if root is null if (root == NULL) { return head; } // First make the sorted linked list // of the left sub-tree head = sortedList(head, root->left); Node* newNode = new Node(root->data); Node* temp = head; Node* prev = NULL; // If linked list is empty add the // node to the head if (temp == NULL) { head = newNode; } else { // Find the correct position of the node // in the given linked list while (temp != NULL) { if (temp->data > root->data) { break; } else { prev = temp; temp = temp->next; } } // Given node is to be attached // at the end of the list if (temp == NULL) { prev->next = newNode; } else { // Given node is to be attached // at the head of the list if (prev == NULL) { newNode->next = temp; head = newNode; } else { // Insertion in between the list newNode->next = temp; prev->next = newNode; } } } // Now add the nodes of the right sub-tree // to the sorted linked list head = sortedList(head, root->right); return head; } // Driver code int main() { /* Tree: 10 / \ 15 2 / \ 1 5 */ treeNode* root = new treeNode(10); root->left = new treeNode(15); root->right = new treeNode(2); root->left->left = new treeNode(1); root->left->right = new treeNode(5); Node* head = sortedList(NULL, root); print(head); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // A linked list node static class Node { int data; Node next; Node(int data) { this.data = data; this.next = null; } }; // A binary tree node static class treeNode { int data; treeNode left; treeNode right; treeNode(int data) { this.data = data; this.left = null; this.right = null; } }; // Function to print the linked list static void print(Node head) { if (head == null) { return; } Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } } // Function to create Linked list from given binary tree static Node sortedList(Node head, treeNode root) { // return head if root is null if (root == null) { return head; } // First make the sorted linked list // of the left sub-tree head = sortedList(head, root.left); Node newNode = new Node(root.data); Node temp = head; Node prev = null; // If linked list is empty add the // node to the head if (temp == null) { head = newNode; } else { // Find the correct position of the node // in the given linked list while (temp != null) { if (temp.data > root.data) { break; } else { prev = temp; temp = temp.next; } } // Given node is to be attached // at the end of the list if (temp == null) { prev.next = newNode; } else { // Given node is to be attached // at the head of the list if (prev == null) { newNode.next = temp; head = newNode; } else { // Insertion in between the list newNode.next = temp; prev.next = newNode; } } } // Now add the nodes of the right sub-tree // to the sorted linked list head = sortedList(head, root.right); return head; } // Driver code public static void main(String[] args) { /* Tree: 10 / \ 15 2 / \ 1 5 */ treeNode root = new treeNode(10); root.left = new treeNode(15); root.right = new treeNode(2); root.left.left = new treeNode(1); root.left.right = new treeNode(5); Node head = sortedList(null, root); print(head); } } // This code is contributed by 29AjayKumar |
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C#
// C# implementation of the approach using System; class GFG { // A linked list node class Node { public int data; public Node next; public Node(int data) { this.data = data; this.next = null; } }; // A binary tree node class treeNode { public int data; public treeNode left; public treeNode right; public treeNode(int data) { this.data = data; this.left = null; this.right = null; } }; // Function to print the linked list static void print(Node head) { if (head == null) { return; } Node temp = head; while (temp != null) { Console.Write(temp.data + " "); temp = temp.next; } } // Function to create Linked list // from given binary tree static Node sortedList(Node head, treeNode root) { // return head if root is null if (root == null) { return head; } // First make the sorted linked list // of the left sub-tree head = sortedList(head, root.left); Node newNode = new Node(root.data); Node temp = head; Node prev = null; // If linked list is empty add the // node to the head if (temp == null) { head = newNode; } else { // Find the correct position of the node // in the given linked list while (temp != null) { if (temp.data > root.data) { break; } else { prev = temp; temp = temp.next; } } // Given node is to be attached // at the end of the list if (temp == null) { prev.next = newNode; } else { // Given node is to be attached // at the head of the list if (prev == null) { newNode.next = temp; head = newNode; } else { // Insertion in between the list newNode.next = temp; prev.next = newNode; } } } // Now add the nodes of the right sub-tree // to the sorted linked list head = sortedList(head, root.right); return head; } // Driver code public static void Main(String[] args) { /* Tree: 10 / \ 15 2 / \ 1 5 */ treeNode root = new treeNode(10); root.left = new treeNode(15); root.right = new treeNode(2); root.left.left = new treeNode(1); root.left.right = new treeNode(5); Node head = sortedList(null, root); print(head); } } // This code is contributed by PrinciRaj1992 |
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Python3
# Python3 implementation of the approach # A linked list node class Node: def __init__(self, data = 0): self.data = data self.next = None # A binary tree node class treeNode: def __init__(self, data): self.data = data self.left = None self.right = None # Function to print the linked list def print_(head): if (head == None): return temp = head while (temp != None): print ( temp.data, end = " " ) temp = temp.next # Function to create Linked list from given binary tree def sortedList( head, root): # return head if root is None if (root == None) : return head # First make the sorted linked list # of the left sub-tree head = sortedList(head, root.left) newNode = Node(root.data) temp = head prev = None # If linked list is empty add the # node to the head if (temp == None) : head = newNode else: # Find the correct position of the node # in the given linked list while (temp != None): if (temp.data > root.data) : break else: prev = temp temp = temp.next # Given node is to be attached # at the end of the list if (temp == None): prev.next = newNode else: # Given node is to be attached # at the head of the list if (prev == None) : newNode.next = temp head = newNode else: # Insertion in between the list newNode.next = temp prev.next = newNode # Now add the nodes of the right sub-tree # to the sorted linked list head = sortedList(head, root.right) return head # Driver code # Tree: # 10 # / \ # 15 2 # / \ #1 5 root = treeNode(10) root.left = treeNode(15) root.right = treeNode(2) root.left.left = treeNode(1) root.left.right = treeNode(5) head = sortedList(None, root) print_(head) # This code is contributed by Arnab Kundu |
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Output:
1 2 5 10 15
Time Complexity: O(n2)
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