Finding the Diameter of an N-Ary Tree: the "To-And-Thru" approach

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Hello folks, this is a problem that exemplifies a technique that I like to call the To-And-Thru. We want to find the diameter of an N-ary tree (meaning a tree where each node has 0, 1 or more children). Take a look: https://leetcode.com/problems/diameter-of-n-ary-tree/

What you want is a near-linear solution in the number of nodes. If you assume that the tree has N nodes, and that each node has an average of LogN children, then the To-And-Thru algorithm below will run in O(N * (LogN)^2) time. Not really linear, but something that seems to be closer to NLogN.

The idea is the following:
  1. Post-Order DFS
  2. For each node, compute the following:
    1. The longest path TO the node
    2. The longest path THRU the node
  3. The above computation will take LogN*LogN since there are two nested loops (I guess there might be a way to reduce this to LogN, meaning one loop)
  4. As you compute 2.1 and 2.2, your diameter will be the Max
See figure below. Works well enough. Code is down below, stay safe, stay motivated!!! ACC



public class Solution {
    public int Diameter(Node root)
    {
        int maxThruNode = 0;
        int maxToNode = 0;
        int diameter = 0;

        Diameter(root, ref maxThruNode, ref maxToNode, ref diameter);

        return diameter > 0 ? diameter - 1 : diameter; //I'm counting the number of nodes in the path. Number of edges is #nodes-1
    }

    private void Diameter(Node node,
                          ref int maxThruNode,
                          ref int maxToNode,
                          ref int diameter)
    {
        if (node == null) return;

        if (node.children.Count == 0)
        {
            maxThruNode = 1;
            maxToNode = 1;
            return;
        }

        int[] childMaxThruNode = new int[node.children.Count];
        int[] childMaxToNode = new int[node.children.Count];

        int index = 0;
        foreach (Node child in node.children)
        {
            Diameter(child, ref childMaxThruNode[index], ref childMaxToNode[index], ref diameter);
            index++;
        }

        maxThruNode = 0;
        maxToNode = 0;
        for (int i = 0; i < childMaxThruNode.Length; i++)
        {
            maxToNode = Math.Max(childMaxToNode[i] + 1, maxToNode);
            diameter = Math.Max(maxToNode, diameter);
            for (int j = i + 1; j < childMaxThruNode.Length; j++)
            {
                maxThruNode = Math.Max(childMaxToNode[i] + childMaxToNode[j] + 1, maxThruNode);
                maxToNode = Math.Max(childMaxToNode[i] + 1, maxToNode);
                maxToNode = Math.Max(childMaxToNode[j] + 1, maxToNode);
                diameter = Math.Max(maxThruNode, diameter);
                diameter = Math.Max(maxToNode, diameter);
            }
        }
    }
}

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