Question

# Hello, I've been working on this for a while and I can't figure the rest out....

Hello, I've been working on this for a while and I can't figure the rest out. Need asap if anyone can help.

maxBSt,minBST,isBST, and inOrder

package lab7;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

public class TreeExercise {
/*
* Construct BST from preorder traversal
*/
public static Node<Integer> consBSTfromPreOrder(int[] arr, int start, int end)
{

if(start > end) return null;

Node<Integer> root = new Node<Integer>(arr[start], null, null);

int index = start+1;
while(index <= end)
{
if (arr[index] < arr[start])
{
index ++;
}
else
{
break;
}
}

root.left = consBSTfromPreOrder(arr, start + 1, index -1);

root.right = consBSTfromPreOrder(arr, index, end);

return root;
}

/*
* Tree level traversal
*/
public static void levelOrder(Node<Integer> root)
{
Queue<Node<Integer>> que = new LinkedList<Node<Integer>>();

que.offer(root);

while(!que.isEmpty())
{

Node<Integer> temp = que.poll();

System.out.print(temp + " ");

if (temp.left != null)
que.offer(temp.left);

if(temp.right != null)
que.offer(temp.right);

}

}

/**
*
* @param root - root of the given BST
* @param ret - an ArrayList of the data following inOrder tree traversal
*/
public static void inOrder(Node<Integer> root, ArrayList<Integer> ret)
{

}

/**
*
* @param root - the root of given BST
* @return the maximum data in the given BST or 0 if empty BST
*/
public static int maxBST(Node<Integer> root)
{
//TODO: please add your code here

return 0; // please remove this line after your coding

}

/**
*
* @param root - the root of given binary search tree
* @return - the minimum data in the given BST or 0 if empty BST
*/

public static int minBST(Node<Integer> root)
{

//TODO: please add your code here

return 0; // please remove this line after your coding
}

/**
*
* @param root - the given root of binary tree
* @return true if a binary search tree; otherwise false
*/

public static boolean isBST(Node<Integer> root)
{
//TODO: please add your code here

return false; // please remove this line after your coding

}

/* ====== Extra Credit (10%) ==============================*/

/**
*
* @param root - the root of the given BST
* @return data following tree level traversal
*/
/*
* Example: 7
* / \
* 5 10
* / \ \
* 3 6 12
*
* return: 7
* 5 10
* 3 6 12
*/
public static ArrayList<ArrayList<Integer>> levelOrderBST(Node<Integer> root)
{

//TODO: add your code here
return null; //remove this line after your coding
}

}

#### Homework Answers

Answer #1
```
/**
*
* @param root - root of the given BST
* @param ret  - an ArrayList of the data following inOrder tree traversal
*/
public static void inOrder(Node<Integer> root, ArrayList<Integer> ret) {
if(root.left != null) {
inOrder(root.left, ret);
}
ret.add(root.data);
if(root.right != null) {
inOrder(root.right, ret);
}
}

/**
*
* @param root - the root of given BST
* @return the maximum data in the given BST or 0 if empty BST
*/
public static int maxBST(Node<Integer> root) {
if(root == null) {
return 0;
}
while(root.right != null) {
root = root.right;
}

return root.data;
}

/**
*
* @param root - the root of given binary search tree
* @return - the minimum data in the given BST or 0 if empty BST
*/

public static int minBST(Node<Integer> root) {

if(root == null) {
return 0;
}
while(root.left != null) {
root = root.left;
}

return root.data;
}

/**
*
* @param root - the given root of binary tree
* @return true if a binary search tree; otherwise false
*/

public static boolean isBST(Node<Integer> root) {
if(root == null) {
return true;
}
if(root.left == null && root.right == null) {
return true;
}

int leftValue = root.data - 1;
int rightValue = root.data + 1;

if(root.left != null) {
leftValue = root.left.data;
}
if(root.right != null) {
rightValue = root.right.data;
}

return (leftValue < root.data) && (rightValue > root.data)
&& isBST(root.left) && isBST(root.right);

}```

please upvote. Thanks!

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