Binary Tree to CDLL

Slug: binary-tree-to-cdll

Difficulty: Medium

Id: 700548

Topic Tags: Linked List, Tree, Data Structures

Company Tags: Amazon, SAP Labs

Summary

Given a binary tree, the task is to convert it into a circular doubly linked list (CDLL). The CDLL should have the same order of nodes as the original binary tree. This problem involves manipulating the tree structure and creating a new linked list.

Detailed Explanation

To solve this problem, we'll first traverse the binary tree in pre-order manner. Then, we'll create a new node for each visited node and link it to the previous node to form a CDLL. The circular part of the CDLL is handled by linking the last node to the first node.

Here's a step-by-step breakdown of the solution:

Initialize two pointers, prev and curr, to NULL.
Traverse the binary tree in pre-order manner (root -> left -> right).
For each visited node:
- Create a new node with the same value as the original node.
- If this is not the first node, link the new node to the previous node (prev).
- Update prev and curr pointers.
Link the last node to the first node to form a circular doubly linked list.

Time Complexity

The time complexity of this solution is O(n), where n is the number of nodes in the binary tree, as we visit each node once.

Space Complexity

The space complexity of this solution is O(n) as well, due to the creation of new nodes and their linking into a CDLL.

Optimized Solutions

Java

class Node { int val; Node left, right; public Node(int val) { this.val = val; this.left = null; this.right = null; } } public class BinaryTreeToCDLL { public static void binaryTreeToCDLL(Node root) { if (root == null) return; Node prev = null, curr = root; while (curr != null) { // Create a new node with the same value as the original node Node newNode = new Node(curr.val); if (prev != null) { // Link the new node to the previous node newNode.right = prev; prev.left = newNode; } curr = curr.right; // Move to the right child prev = newNode; // Update the previous node } // Link the last node to the first node to form a circular doubly linked list prev.right = root; root.left = prev; } }