Finding last element of a binary heap

Question

quoting Wikipedia:

It is perfectly acceptable to use a traditional binary tree data structure to implement a binary heap. There is an

Answer 1

Solution in case you don't have reference to parent !!! To find the right place for next node you have 3 cases to handle

• case (1) Tree level is complete Log2(N)
• case (2) Tree node count is even
• case (3) Tree node count is odd

Insert:

void Insert(Node root,Node n)
{
Node parent = findRequiredParentToInsertNewNode (root);
if(parent.left == null)
parent.left = n;
else
parent.right = n;
}

Find the parent of the node in order to insert it

void findRequiredParentToInsertNewNode(Node root){

Node last = findLastNode(root);

 //Case 1 
 if(2*Math.Pow(levelNumber) == NodeCount){
     while(root.left != null)
      root=root.left;
  return root;
 }
 //Case 2
 else if(Even(N)){
   Node n =findParentOfLastNode(root ,findParentOfLastNode(root ,last));
 return n.right;
 }
//Case 3
 else if(Odd(N)){
  Node n =findParentOfLastNode(root ,last);
 return n;
 }

}

To find the last node you need to perform a BFS (breadth first search) and get the last element in the queue

 Node findLastNode(Node root)
 {
     if (root.left == nil)
         return root

     Queue q = new Queue();

     q.enqueue(root);
     Node n = null;

     while(!q.isEmpty()){
      n = q.dequeue();
     if ( n.left != null )
         q.enqueue(n.left);
     if ( n.right != null )
         q.enqueue(n.right);
        }
   return n;
}

Find the parent of the last node in order to set the node to null in case replacing with the root in removal case

 Node findParentOfLastNode(Node root ,Node lastNode)
{
    if(root == null)
        return root;

    if( root.left == lastNode || root.right == lastNode )
        return root;

    Node n1= findParentOfLastNode(root.left,lastNode);
    Node n2= findParentOfLastNode(root.left,lastNode);

    return n1 != null ? n1 : n2;
}