Diameter of N-Ary Tree

Problem Statement

In this task, you are presented with the root of an N-ary tree, and your goal is to determine the diameter of this tree. The diameter is defined as the longest possible path between any two nodes within the tree. It's important to note that this path might not necessarily pass through the root. The tree is structured in N-ary format, where each node can have multiple children (not just two as in binary trees), and the tree data is typically represented by a level-order traversal, where each group of children is concluded with a null value to indicate moving to the next level.

Examples

Example 1

Input:

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

Output:

3

Explanation:

Diameter is shown in red color.

Example 2

Input:

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

Output:

4

Example 3

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:

7

Constraints

• The depth of the n-ary tree is less than or equal to 1000.
• The total number of nodes is between [1, 104].

Approach and Intuition

To approach the problem of finding the diameter of an N-ary tree, we can draw intuition from methods used in binary tree diameter calculations, but we must adjust for the fact that any node can have multiple children. Here's a step-by-step breakdown of a possible approach:

1. Use Depth-First Search (DFS) to explore the tree. The key in DFS for this problem is to find out the maximum two depths you can achieve from any node because the potential longest path will be between these two deepest subtrees originating from the same node.
2. Initiate DFS from the root. For each node, compute the depths of its subtrees.
3. At each node during DFS traversal, calculate:
   • The deepest path length ending at this node.
   • The second deepest path length also ending at this node.
4. The potential diameter at each node will be the sum of its two longest child paths (because forming a path using these two longest paths will potentially give a path longer than passing through the root or any other nodes).
5. As you perform DFS and bubble up back to the root, keep track of the global maximum diameter found at any node in the tree.
6. The result when the DFS completes will be this global maximum value, which represents the diameter of the tree.

The constraints that the tree depth is less than or equal to 1000 and the total number of nodes is between 1 and 10,000 ensure that this depth-first search approach will execute efficiently within practical time limits.

Solutions

class Solution {
    protected int maxDiameter = 0;
    
    protected int depthCalc(Node node, int depth) {
        if (node.children.size() == 0)
            return depth;
    
        int deepest = depth, secondDeepest = 0;
        for (Node child : node.children) {
            int childDepth = depthCalc(child, depth + 1);
            if (childDepth > deepest) {
                secondDeepest = deepest;
                deepest = childDepth;
            } else if (childDepth > secondDeepest) {
                secondDeepest = childDepth;
            }
            int currentDistance = deepest + secondDeepest - 2 * depth;
            this.maxDiameter = Math.max(this.maxDiameter, currentDistance);
        }
    
        return deepest;
    }
    
    public int getDiameter(Node root) {
        this.maxDiameter = 0;
        depthCalc(root, 0);
        return maxDiameter;
    }
}