10.1 BinaryHeap: An Implicit Binary Tree

Our first implementation of a (priority) Queue is based on a technique that is over four hundred years old. Eytzinger's method allows us to represent a complete binary tree as an array by laying out the nodes of the tree in breadth-first order (see Section 6.1.2). In this way, the root is stored at position 0, the root's left child is stored at position 1, the root's right child at position 2, the left child of the left child of the root is stored at position 3, and so on. See Figure 10.1.

If we apply Eytzinger's method to a sufficiently large tree, some
patterns emerge. The left child of the node at index
is at index
and the right child of the node at index
is at
index
. The parent of the node at index
is at
index
.

A BinaryHeap uses this technique to implicitly represent a complete binary tree in which the elements are heap-ordered: The value stored at any index is not smaller than the value stored at index , with the exception of the root value, . It follows that the smallest value in the priority Queue is therefore stored at position 0 (the root).

In a BinaryHeap, the
elements are stored in an array
:

Implementing the
operation is fairly straightforward. As with
all array-based structures, we first check to see if
is full (by checking if
) and, if so, we grow
. Next, we place
at location
and increment
. At this point, all that remains is to ensure
that we maintain the heap property. We do this by repeatedly swapping
with its parent until
is no longer smaller than its parent.
See Figure 10.2.

Implementing the operation, which removes the smallest value from the heap, is a little trickier. We know where the smallest value is (at the root), but we need to replace it after we remove it and ensure that we maintain the heap property.

The easiest way to do this is to replace the root with the value
, delete
that value, and decrement
. Unfortunately, the new root element is now
probably not the smallest element, so it needs to be moved downwards.
We do this by repeatedly comparing this element to its two children.
If it is the smallest of the three then we are done. Otherwise, we swap
this element with the smallest of its two children and continue.

As with other array-based structures, we will ignore the time spent in calls to , since these can be accounted for using the amortization argument from Lemma 2.1. The running times of both and then depend on the height of the (implicit) binary tree. Luckily, this is a complete binary tree; every level except the last has the maximum possible number of nodes. Therefore, if the height of this tree is , then it has at least nodes. Stated another way

The following theorem summarizes the performance of a BinaryHeap:

*Furthermore, beginning with an empty BinaryHeap, any sequence of
and
operations results in a total of
time spent during all calls to
.*

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