## Code Snippet

package com.ideserve.questions.nilesh;
/**
* <b>IDeserve <br>
* <a href="https://www.youtube.com/c/IDeserve">https://www.youtube.com/c/IDeserve</a>
* Checks if identical BSTs without building them in time O(n^2).
* @author Nilesh
*/
public class CheckIdenticalBSTs
{
private boolean checkIfSameBSTs(int[] arrayForTree1, int[] arrayForTree2, int index1, int index2, int min, int max)
{
/*
* find the first element between min and max.
* that element would be used as the root of the subtree we are looking to construct
*/
int i,j;
for (i = index1; i < arrayForTree1.length; i++)
{
if ((min < arrayForTree1[i]) && (arrayForTree1[i] < max))
{
break;
}
}
for (j = index2; j < arrayForTree2.length; j++)
{
if ((min < arrayForTree2[j]) && (arrayForTree2[j] < max))
{
break;
}
}
/*
* since we are constructing same part of the trees for both arrays
* element found should be same. If element not found,
* then that means we found no node satisfying this condition.
* If this is the case then this should be the case for both the trees
*/
if ((i == arrayForTree1.length) && (j == arrayForTree2.length))
{
return true; // no node found for both trees
}
if ((i == arrayForTree1.length) || (j == arrayForTree2.length))
{
return false; // no node found only in case of one of the trees
}
// if current nodes are same then check if left and right sub-trees for current nodes are same
if (arrayForTree1[i] == arrayForTree2[j])
{
return (checkIfSameBSTs(arrayForTree1, arrayForTree2, i+1, j+1, min, arrayForTree1[i]) &&
checkIfSameBSTs(arrayForTree1, arrayForTree2, i+1, j+1, arrayForTree1[i], max));
}
return false;
}
public boolean checkIfSameBSTs(int[] arrayForTree1, int[] arrayForTree2)
{
return checkIfSameBSTs(arrayForTree1, arrayForTree2, 0, 0, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
public static void main(String[] args)
{
CheckIdenticalBSTs solution = new CheckIdenticalBSTs();
int[] arrayForTree1 = {3,5,4,6,1,0,2};
int[] arrayForTree2 = {3,1,5,2,4,6,0};
System.out.println("Check if identical BSTs: " + solution.checkIfSameBSTs(arrayForTree1, arrayForTree2));
}
}