Clone N-ary Tree

Problem Statement

In this task, we are working with an N-ary tree where each node can have multiple children. The goal is to make a deep copy of the given N-ary tree. A deep copy implies creating a new instance of the tree with the same structure and values as the original, but completely independent of it in memory.

Each node in the N-ary tree has an integer value and a list of its children nodes. Serialization of this tree is defined by the level-order traversal, where groups of a node's children are distinguished by a 'null' value to denote transitions between different levels or siblings.

Our objective is to handle this copying mechanism correctly to ensure that the new tree reflects the exact same structure and values as the original but operates as a separate entity.

Examples

Example 1

Input:

root = [1,null,3,2,4,null,5,6]

Output:

[1,null,3,2,4,null,5,6]

Example 2

Input:

root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Output:

[1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Constraints

• The depth of the n-ary tree is less than or equal to 1000.
• The total number of nodes is between [0, 104].

Approach and Intuition

The problem involves creating a duplicate of an N-ary tree while maintaining the hierarchical structure and values of the original tree. Here, we focus on understanding the traversal and copying method needed:

1. Utilize a traversal technique that helps in visiting each node and copying its values and structure. Level-order (Breadth-First Search) traversal is a natural fit for this, given the way the tree is serialized.

2. For each node encountered:

   • Create a new node with the same value.
   • Recursively copy all its children by applying the same cloning method to each child. This ensures a deep copy where changes to the original tree's nodes or subtrees do not affect the cloned tree.

3. Given the constraints, make sure the solution accommodates up to 10,000 nodes and handles a maximum depth of 1000. This requires careful management of recursion depth and memory usage, specifically ensuring that the system does not run into a stack overflow or excessive memory consumption.

From the examples provided:

• Example 1 shows a simple tree structure and its direct copy. This helps verify that the basic copying mechanism works for trees with a few nodes and levels.
• Example 2, with a more complex and deeper structure, tests the algorithm's ability to handle larger trees with more levels and children. This ensures that the copying mechanism correctly handles extensive and nested child lists.

Both examples illustrate the unchanged structural integrity and values in the copied trees, verifying that the solution should perform a deep copy without unintentionally linking parts of the new tree to the original. Thus, an efficiently implemented recursive or iterative copying mechanism using a queue (for iterative deep copying using BFS) or system stack (for recursive DFS) is essential.

Solutions

class Solution {
    public Node copyTree(Node original) {
        if (original == null) {
            return original;
        }

        Node copiedRoot = new Node(original.val);
        
        Deque<Node[]> trackingQueue = new ArrayDeque<>();
        trackingQueue.addLast(new Node[]{original, copiedRoot});

        while (!trackingQueue.isEmpty()) {
            Node[] currentPair = trackingQueue.removeFirst();

            Node sourceNode = currentPair[0];
            Node targetNode = currentPair[1];
            for (Node child : sourceNode.children) {
                Node copiedChild = new Node(child.val);

                targetNode.children.add(copiedChild);
                trackingQueue.addLast(new Node[]{child, copiedChild});
            }
        }

        return copiedRoot;
    }
}

class Solution:
    def duplicateTree(self, root: 'Node') -> 'Node':

        if not root:
            return root

        copied_root = Node(root.val)
        # Initialize the queue for BFS traversal.
        bfs_queue = deque([(root, copied_root)])

        while bfs_queue:
            # Extract nodes from the front of the queue.
            source, clone = bfs_queue.popleft()

            for child in source.children:
                copy_child = Node(child.val)

                # Append the copied children to the clone's children list.
                clone.children.append(copy_child)

                # Add to queue for further processing.
                bfs_queue.append((child, copy_child))

        return copied_root