1522. Diameter of N-Ary Tree

Given a root of an N-ary tree, you need to compute the length of the diameter of the tree.

The diameter of an N-ary tree is the length of the longest path between any two nodes in the tree. This path may or may not pass through the root.

(Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value.)

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].

Solution:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;


    public Node() {
        children = new ArrayList<Node>();
    }

    public Node(int _val) {
        val = _val;
        children = new ArrayList<Node>();
    }

    public Node(int _val,ArrayList<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    int result = 0;
    public int diameter(Node root) {
        if (root.children.size() == 0){
            return 0;
        }
        height(root);
        return result;
    }

    public int height(Node root){

        if (root == null){
            return 0;
        }

        PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
        for (Node child : root.children){
            int childHeight = height(child);
            pq.offer(childHeight);
            if (pq.size() > 2){
                pq.poll();
            }
        }

        if (pq.isEmpty()){
            result = Math.max(result, 1);
            return 1;
        }else if (pq.size() == 1){
            int one = pq.poll();
            result = Math.max(result, one);
            return one + 1;

        }else{
            int two = pq.poll();
            int one = pq.poll();
            result = Math.max(result, one + two);

            return one + 1;
        }

    }
}

// TC: O(nlogn)
// SC: O(n)