Lowest Common Ancestor (generic Tree)

I/P:

24 10 20 50 -1 60 -1 -1 30 70 -1 80 110 -1 120 -1 -1 90 -1 -1 40 100 -1 -1 -1 120 80

O/P: 80

Store node-to-root path of the first node in array p1. 

Store node-to-root path of the second node in array p2. 

Initialize two pointers i and j as p1.length - 1 and p2.length - 1 respectively. 

Decrement i and j until arr[i] and arr[j] become unequal. 

The lowest common ancestor will be arr[i + 1] (or arr[j + 1]).

package pep.Day29;

import java.io.*;
import java.util.*;

public class Lowest_Common_Ancestor_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 ArrayList<Integer> nodeToRootPath(Node node, int data) {
        if (node.data == data) {
            ArrayList<Integer> base = new ArrayList<>();
            base.add(data);
            return base;
        }

        for (Node child : node.children) {
            ArrayList<Integer> pathTillChild = nodeToRootPath(child, data);
            if (pathTillChild.size() > 0) {
                pathTillChild.add(child.data);
                return pathTillChild;
            }
        }

        return new ArrayList<Integer>();
    }

    public static int lca(Node node, int d1, int d2) {
        ArrayList<Integer> first = nodeToRootPath(node, d1);
        ArrayList<Integer> second = nodeToRootPath(node, d2);

        int i = first.size() - 1, j = second.size() - 1;
        while (i >= 0 && j >= 0 && first.get(i) == second.get(j)) {
            i--;
            j--;
        }
        return first.get(i + 1);

    }

    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]);
        }

        int d1 = Integer.parseInt(br.readLine());
        int d2 = Integer.parseInt(br.readLine());

        Node root = construct(arr);
        int lca = lca(root, d1, d2);
        System.out.println(lca);
        // display(root);
    }

}

Time Complexity: O(n) Finding the node in the entire tree to get it's node to the root path is an O(n) task. Then, just traversing the node-to-root path (arrays) takes O(d) where d = depth of node. In the worst case, d can be equal to n, hence total time complexity will be O(n) only.

Space Complexity: We are storing node-to-root paths in arraylists. This will take O(n) auxiliary space.