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Microsoft Interview Question for Interns


Country: United States
More Questions from This Interview



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of 0 vote

what is an unordered tree?

- veeru February 09, 2017 | Flag Reply
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class Node
	{
		int data;
		Node left, right;
		
		public Node(int key)
		{
			data = key;
			left = null;
			right = null;
		}
	}
	
	class unordered 
	{
		Node root;
		
		unordered(int key)
		{
			root = new Node(key);
		}
		unordered()
		{
			root=null;
		}
		
		public void uotollrec(Node roo, List<Integer> ll)
		{
			
			if(roo==null)
				return;
			ll.add(roo.data);
			uotollrec(roo.left,ll);
			uotollrec(roo.right,ll);
			
		}
	}
	
	class BinarySearchTree
	{
		Node root;
		
		BinarySearchTree() {
			root = null;
		}
		
		void addbinary(int item)
		{
			root = addrec(root,item);
		}
		
		Node addrec(Node root, int item)
		{
			if(root == null)
			{
				root = new Node(item);
				return root;
			}
			if (root.data>item)
				root.left=addrec(root.left,item);
			else
				root.right=addrec(root.right,item);
			
			return root;
		}
		
		void trav()
		{
			traverse1(root);
		}
		void traverse1(Node roo)
		{
			
			if(roo!=null)
			{
			System.out.println(roo.data);
			traverse1(roo.left);
			traverse1(roo.right);
			}
			
		}
	}
	
	public class first
	{
		static List<Integer> ll = new ArrayList<Integer>();
		BinarySearchTree bst = new BinarySearchTree();
		
		public void uotobinary(unordered uo)
		{
			int r;
			unordered u1 = new unordered();
			u1.uotollrec(uo.root,ll);
			Collections.sort(ll);
			if(ll.size()%2==0)
				r = ll.size()/2-1 ;
			else
				r = (ll.size()/2);
			bst.addbinary(ll.get(r));
			
			for(int i = r-1; i>=00;i--)
				bst.addbinary(ll.get(i));
			
			for(int i = r+1; i<ll.size() ; i++)
				bst.addbinary(ll.get(i));
			
			bst.trav();
			
		}	
	
	public static void main(String[] args) 
    {
		first f1 = new first();
    	unordered uo = new unordered();
   		uo.root = new Node(7);
   		uo.root.left=new Node(6);
   		uo.root.right=new Node(5);
   		uo.root.left.left=new Node(4);
   		uo.root.left.right=new Node(3);
   		uo.root.right.left=new Node(2);
   		uo.root.right.right=new Node(1);   		
   		
   		f1.uotobinary(uo);   		
    }
}

- PalakMundra March 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Node
	{
		int data;
		Node left, right;
		
		public Node(int key)
		{
			data = key;
			left = null;
			right = null;
		}
	}
	
	class unordered 
	{
		Node root;
		
		unordered(int key)
		{
			root = new Node(key);
		}
		unordered()
		{
			root=null;
		}
		
		public void uotollrec(Node roo, List<Integer> ll)
		{
			
			if(roo==null)
				return;
			ll.add(roo.data);
			uotollrec(roo.left,ll);
			uotollrec(roo.right,ll);
			
		}
	}
	
	class BinarySearchTree
	{
		Node root;
		
		BinarySearchTree() {
			root = null;
		}
		
		void addbinary(int item)
		{
			root = addrec(root,item);
		}
		
		Node addrec(Node root, int item)
		{
			if(root == null)
			{
				root = new Node(item);
				return root;
			}
			if (root.data>item)
				root.left=addrec(root.left,item);
			else
				root.right=addrec(root.right,item);
			
			return root;
		}
		
		void trav()
		{
			traverse1(root);
		}
		void traverse1(Node roo)
		{
			
			if(roo!=null)
			{
			System.out.println(roo.data);
			traverse1(roo.left);
			traverse1(roo.right);
			}
			
		}
	}
	
	public class first
	{
		static List<Integer> ll = new ArrayList<Integer>();
		BinarySearchTree bst = new BinarySearchTree();
		
		public void uotobinary(unordered uo)
		{
			int r;
			unordered u1 = new unordered();
			u1.uotollrec(uo.root,ll);
			Collections.sort(ll);
			if(ll.size()%2==0)
				r = ll.size()/2-1 ;
			else
				r = (ll.size()/2);
			bst.addbinary(ll.get(r));
			
			for(int i = r-1; i>=00;i--)
				bst.addbinary(ll.get(i));
			
			for(int i = r+1; i<ll.size() ; i++)
				bst.addbinary(ll.get(i));
			
			bst.trav();
			
		}	
	
	public static void main(String[] args) 
    {
		first f1 = new first();
    	unordered uo = new unordered();
   		uo.root = new Node(7);
   		uo.root.left=new Node(6);
   		uo.root.right=new Node(5);
   		uo.root.left.left=new Node(4);
   		uo.root.left.right=new Node(3);
   		uo.root.right.left=new Node(2);
   		uo.root.right.right=new Node(1);   		
   		
   		f1.uotobinary(uo);   		
    }
}

- PalakMundra March 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Node
	{
		int data;
		Node left, right;
		
		public Node(int key)
		{
			data = key;
			left = null;
			right = null;
		}
	}
	
	class unordered 
	{
		Node root;
		
		unordered(int key)
		{
			root = new Node(key);
		}
		unordered()
		{
			root=null;
		}
		
		public void uotollrec(Node roo, List<Integer> ll)
		{
			
			if(roo==null)
				return;
			ll.add(roo.data);
			uotollrec(roo.left,ll);
			uotollrec(roo.right,ll);
			
		}
	}
	
	class BinarySearchTree
	{
		Node root;
		
		BinarySearchTree() {
			root = null;
		}
		
		void addbinary(int item)
		{
			root = addrec(root,item);
		}
		
		Node addrec(Node root, int item)
		{
			if(root == null)
			{
				root = new Node(item);
				return root;
			}
			if (root.data>item)
				root.left=addrec(root.left,item);
			else
				root.right=addrec(root.right,item);
			
			return root;
		}
		
		void trav()
		{
			traverse1(root);
		}
		void traverse1(Node roo)
		{
			
			if(roo!=null)
			{
			System.out.println(roo.data);
			traverse1(roo.left);
			traverse1(roo.right);
			}
			
		}
	}
	
	public class first
	{
		static List<Integer> ll = new ArrayList<Integer>();
		BinarySearchTree bst = new BinarySearchTree();
		
		public void uotobinary(unordered uo)
		{
			int r;
			unordered u1 = new unordered();
			u1.uotollrec(uo.root,ll);
			Collections.sort(ll);
			if(ll.size()%2==0)
				r = ll.size()/2-1 ;
			else
				r = (ll.size()/2);
			bst.addbinary(ll.get(r));
			
			for(int i = r-1; i>=00;i--)
				bst.addbinary(ll.get(i));
			
			for(int i = r+1; i<ll.size() ; i++)
				bst.addbinary(ll.get(i));
			
			bst.trav();
			
		}	
	
	public static void main(String[] args) 
    {
		first f1 = new first();
    	unordered uo = new unordered();
   		uo.root = new Node(7);
   		uo.root.left=new Node(6);
   		uo.root.right=new Node(5);
   		uo.root.left.left=new Node(4);
   		uo.root.left.right=new Node(3);
   		uo.root.right.left=new Node(2);
   		uo.root.right.right=new Node(1);   		
   		
   		f1.uotobinary(uo);   		
    }
}

- PalakMundra March 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Node
	{
		int data;
		Node left, right;
		
		public Node(int key)
		{
			data = key;
			left = null;
			right = null;
		}
	}
	
	class unordered 
	{
		Node root;
		
		unordered(int key)
		{
			root = new Node(key);
		}
		unordered()
		{
			root=null;
		}
		
		public void uotollrec(Node roo, List<Integer> ll)
		{
			
			if(roo==null)
				return;
			ll.add(roo.data);
			uotollrec(roo.left,ll);
			uotollrec(roo.right,ll);
			
		}
	}
	
	class BinarySearchTree
	{
		Node root;
		
		BinarySearchTree() {
			root = null;
		}
		
		void addbinary(int item)
		{
			root = addrec(root,item);
		}
		
		Node addrec(Node root, int item)
		{
			if(root == null)
			{
				root = new Node(item);
				return root;
			}
			if (root.data>item)
				root.left=addrec(root.left,item);
			else
				root.right=addrec(root.right,item);
			
			return root;
		}
		
		void trav()
		{
			traverse1(root);
		}
		void traverse1(Node roo)
		{
			
			if(roo!=null)
			{
			System.out.println(roo.data);
			traverse1(roo.left);
			traverse1(roo.right);
			}
			
		}
	}
	
	public class first
	{
		static List<Integer> ll = new ArrayList<Integer>();
		BinarySearchTree bst = new BinarySearchTree();
		
		public void uotobinary(unordered uo)
		{
			int r;
			unordered u1 = new unordered();
			u1.uotollrec(uo.root,ll);
			Collections.sort(ll);
			if(ll.size()%2==0)
				r = ll.size()/2-1 ;
			else
				r = (ll.size()/2);
			bst.addbinary(ll.get(r));
			
			for(int i = r-1; i>=00;i--)
				bst.addbinary(ll.get(i));
			
			for(int i = r+1; i<ll.size() ; i++)
				bst.addbinary(ll.get(i));
			
			bst.trav();
			
		}	
	
	public static void main(String[] args) 
    {
		first f1 = new first();
    	unordered uo = new unordered();
   		uo.root = new Node(7);
   		uo.root.left=new Node(6);
   		uo.root.right=new Node(5);
   		uo.root.left.left=new Node(4);
   		uo.root.left.right=new Node(3);
   		uo.root.right.left=new Node(2);
   		uo.root.right.right=new Node(1);   		
   		
   		f1.uotobinary(uo);   		
    }
}

- Palak.1009 March 28, 2018 | Flag Reply


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