Question

in java please thanks!

Given a pointer to a binary tree write a routine which will traverse the tree and return the number of nodes in the tree whose information field which is an integer is between 59 and 112 In addition return the value of the node with the largest integer and as well return a count of the number of nodes that have no sons. For the 3rd part do not include nodes that have both right and left sons as well as do not include any nodes that have only a right or nodes that just have a left IV. son.

Answer 1

Node.java

public class Node {
public int data;
public Node left, right;
  
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
}

BinaryTree.java

public class BinaryTree {
  
public Node root;
private static int count = 0;
  
public BinaryTree()
{
this.root = null;
}

public Node getRoot() {
return root;
}

public void setRoot(Node root) {
this.root = root;
}
  
public int searchBinaryTree(Node temp)
{
if(temp == null)
return 0;
else
{
if(temp.data >= 59 && temp.data <= 112)
return (1 + this.searchBinaryTree(temp.left) + this.searchBinaryTree(temp.right));
}
return this.searchBinaryTree(temp.right) + this.searchBinaryTree(temp.left);
}
  
public int findMaximumNodeValue(Node temp)
{
if(temp == null)
return Integer.MIN_VALUE;
  
int result = temp.data;
int leftResult = findMaximumNodeValue(temp.left);
int rightResult = findMaximumNodeValue(temp.right);
  
if(leftResult > result)
result = leftResult;
if(rightResult > result)
result = rightResult;
return result;
}
  
public int countOfNodesHavingNoSons(Node temp)
{
if(temp == null || (temp.left == null && temp.right == null))
return 0;
  
return 1 + countOfNodesHavingNoSons(temp.left) + countOfNodesHavingNoSons(temp.right);
}
}

BinaryTreeDemo.java (Main class)

public class BinaryTreeDemo {
  
public static void main(String[]args)
{
BinaryTree bt1 = new BinaryTree();
  
bt1.root = new Node(60);
bt1.root.left = new Node(69);
bt1.root.right = new Node(59);
bt1.root.left.left = new Node(71);
bt1.root.left.right = new Node(21);
bt1.root.right.left = new Node(83);
bt1.root.right.right = new Node(100);
  
System.out.println("Number of nodes having value between 59 and 112 is = " + bt1.searchBinaryTree(bt1.root));
System.out.println("Node having maximum value is = " + bt1.findMaximumNodeValue(bt1.root));
  
BinaryTree bt2 = new BinaryTree();
bt2.root = new Node(33);
bt2.root.left = new Node(45);
bt2.root.right = new Node(25);
bt2.root.left.left = new Node(61);
bt2.root.left.right = new Node(17);
  
System.out.println("Count of nodes having no sons is = " + bt2.countOfNodesHavingNoSons(bt2.root) + "\n");
}
}

***************************************************************** SCREENSHOT **********************************************************

DD run Number of nodes having value between 59 and 112 is= 6 Node having maximum value is =100 Count of nodes having no sons is= 2 BUILD SUCCESSFUL (total time: 0 seconds)