Tuesday, March 1, 2022

Are Trees Similar In Shape

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 24 1 2 5 -1 6 -1 -1 3 7 -1 8 11 -1 12 -1 -1 9 -1 -1 4 10 -1 -1 -1


O/P: true


1. first verify the size of children of both the nodes

2. if same, then tell their children to check their children further.



package pep.Day30;

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

public class Are_Trees_Similar_In_Shape {

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 boolean areSimilar(Node n1, Node n2) {
// write your code here
if (n1.children.size() != n2.children.size())
return false;
else {
for (int i = 0; i < n1.children.size(); i++) {
boolean bool = areSimilar(n1.children.get(i), n2.children.get(i));
if (!bool)
return false;
}
}

return true;
}

public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

int n1 = Integer.parseInt(br.readLine());
int[] arr1 = new int[n1];
String[] values1 = br.readLine().split(" ");
for (int i = 0; i < n1; i++) {
arr1[i] = Integer.parseInt(values1[i]);
}
Node root1 = construct(arr1);

int n2 = Integer.parseInt(br.readLine());
int[] arr2 = new int[n2];
String[] values2 = br.readLine().split(" ");
for (int i = 0; i < n2; i++) {
arr2[i] = Integer.parseInt(values2[i]);
}
Node root2 = construct(arr2);

boolean similar = areSimilar(root1, root2);
System.out.println(similar);
}

}



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