Find Root of N-Ary Tree

Problem Statement

In the given problem, you are entrusted with a seemingly straightforward but intriguing task: determining the root of an N-ary tree from an array of its nodes. Each node, represented as a Node object, carries a unique value which eliminates the possibility of duplicate values within the tree. The unique challenge presented here lies in identifying the root node without the explicit structure of the tree, since the nodes are provided in an arbitrary sequence.

To clarify the serialization method involved, the N-ary tree's structure is captured through a level order traversal, where breaks in continuity of child nodes are marked with the null value. This serialized form is crucial as it serves both as a means for reconstructing the tree in tests and validating the correctness of your function's output.

During tests, your function is expected to find and return the root node based on the provided unordered array of nodes. The system that tests this functionality will serialize the node returned by your function and match it against the originally provided serialized data to ascertain the accuracy of the solution.

Examples

Example 1

Input:

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

Output:

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

Explanation:

The tree from the input data is shown above.
The driver code creates the tree and gives findRoot the Node objects in an arbitrary order.
For example, the passed array could be [Node(5),Node(4),Node(3),Node(6),Node(2),Node(1)] or [Node(2),Node(6),Node(1),Node(3),Node(5),Node(4)].
The findRoot function should return the root Node(1), and the driver code will serialize it and compare with the input data.
The input data and serialized Node(1) are the same, so the test passes.

Example 2

Input:

tree = [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 total number of nodes is between [1, 5 * 104].
• Each node has a unique value.

Approach and Intuition

To work towards finding the root of the N-ary tree based on the given constraints and examples, consider the following steps informed by the intuition about tree structures and node relationships:

1. Understand the Properties of Root Node:

   • The root node is the only node in the tree without any parent.
   • Every other node in the tree will have exactly one parent.

2. Initial Thoughts on Implementation:

   • Given an array of nodes, each of which knows its children but not its parent, the challenge becomes identifying the node which is not listed as a child of any other node.
   • If you can demonstrate that a node does NOT appear as a child of any other node in the array, this node must be the root.

3. Potential Approach:

   • Create a set or another suitable data structure to track all the child nodes seen as you iterate through the array of Node objects.
   • Iterate through each node's children and add them to this set.
   • Once all nodes have been processed, any node not in your "children set" is the root.

4. Optimization Considerations:

   • Given the constraints of nodes (between 1 and 50,000), the approach to process each node in a single pass (O(n) complexity regarding the number of nodes) would be efficient.
   • Careful attention must be paid to the memory usage, especially the structure used to store child references. A balance between time complexity and spatial complexity is vital.

This approach, rooted in fundamental properties of trees, offers a straightforward solution to identifying the root node without reconstructing the entire tree, which ensures efficiency given the constraints on the number of nodes and their unique values.

Solutions

class Solution {
    public Node locateRootNode(List<Node> treeNodes) {

        Integer accumulatedValue = 0;

        for (Node node : treeNodes) {
            accumulatedValue += node.val;
            for (Node childNode : node.children)
                accumulatedValue -= childNode.val;
        }

        Node rootNode = null;
        for (Node node : treeNodes) {
            if (node.val == accumulatedValue) {
                rootNode = node;
                break;
            }
        }
        return rootNode;
    }
}

"""
# Class structure for a Tree Node.
class TreeNode:
    def __init__(self, value=None, kids=None):
        self.value = value
        self.kids = kids if kids is not None else []
"""

class SolutionFinder:
    def locateRootNode(self, nodeList: List['TreeNode']) -> 'TreeNode':
        accumulated_value = 0

        for treeNode in nodeList:
            # Adding value from a potential root candidate
            accumulated_value += treeNode.value
            for kid in treeNode.kids:
                # Removing value from a non-root candidate
                accumulated_value -= kid.value

        # Identifying root node based on accumulated sum
        for treeNode in nodeList:
            if treeNode.value == accumulated_value:
                return treeNode