Implementing a Heap/Priority Queueu

• A complete binary tree
  • Remember the binary tree representation in an array
  • The problem was, if we had all left children, this is not the case whith a heap.
  • But we do have easy access to where to insert next!
  • Convince yourself of this.
  • n = number of nodes in the tree
  • n = last position in the array used.
• Inserting an item into the heap
  • We must put it at the next available space.
  • Then fix the heap
  • Do this by fixing the heap from the new leaf up to the root
  • Let the value bubble up.

    Bubble UP (newchild)
    (1) if (!isroot(newchild))
    (2)   if newchild &k newchild.parent()
    (3)       swap newchild.parent, newchild
    (4)        bubbleup(newchild)
    	

    • O(h), h = height,
    • But height is O(lg n)
• Extracting the min from the heap:
  • remove the value at the root.
  • Place the value at the farthest node in the root.
  • Bubble the new value down

  • Bubble Down(node) 
    (1)   if !isleaf(node) 
    (2)      if node.rightChild.isExternal()
    (3)         s = rightchild
    (4)      else 
    (5)         s = min(leftchild.element,rightchild.element)
    (6)      if s.element < node.element
    (7)         swap(node,s)
    (8)	 BubbleDown(node)
           

• Again, this is O(h) = O(lg(n))