Height Of Generic Tree

I/P: 

12

10 20 -1 30 50 -1 60 -1 -1 40 -1 -1

O/P: 2

For this we look at case 1 where a tree has only one element, in such a case height is 0 (in terms of edges). Therefore we will initialize our height variable with -1, so that when it does self work, one gets added to this making height of such a case 0.

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Stack;

public class Height_Of_GenericTree {
    private static class Node {
        int data;
        ArrayList<Node> children = new ArrayList<>();
    }

    public static void display(Node node) {
        String str = node.data + " -> ";
        for (Node child : node.children) {
            str += child.data + ", ";
        }
        str += ".";
        System.out.println(str);

        for (Node child : node.children) {
            display(child);
        }
    }

    public static Node construct(int[] arr) {
        Node root = null;

        Stack<Node> st = new Stack<>();
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == -1) {
                st.pop();
            } else {
                Node t = new Node();
                t.data = arr[i];

                if (st.size() > 0) {
                    st.peek().children.add(t);
                } else {
                    root = t;
                }

                st.push(t);
            }
        }

        return root;
    }

    public static int height(Node node) {
        // why height initiazed to -1 ??
        // if I'm at root node, height should be 0
        // but since, at the end we are incrementing height by 1
        // Hence, height should be initialized to -1.
        int height = -1;
        for (Node child : node.children) {
            height = Math.max(height, height(child));
        }
        height += 1;
        return height;
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int[] arr = new int[n];
        String[] values = br.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(values[i]);
        }

        Node root = construct(arr);
        int h = height(root);
        System.out.println(h);
        // display(root);
    }

}

Time Complexity:

O(n) We travel to all the elements of the tree. Every node is processed individually inside the 'for' loop.

Space Complexity:

O(1) No extra space is used. But when the function runs at that time, space acquired by the memory stack will be O(n).