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6.1 Heaps


What are the minimum and maximum numbers of elements in a heap of height $h$?

At least $2^h$ and at most $2^{h + 1} − 1$. Can be seen because a complete binary tree of depth $h − 1$ has $\sum_{i = 0}^{h - 1} 2^i = 2^h - 1$ elements, and the number of elements in a heap of depth $h$ is between the number for a complete binary tree of depth $h − 1$ exclusive and the number in a complete binary tree of depth $h$ inclusive.


Show that an $n$-element heap has height $\lfloor \lg n \rfloor$.

Write $n = 2^m − 1 + k$ where $m$ is as large as possible. Then the heap consists of a complete binary tree of height $m − 1$, along with $k$ additional leaves along the bottom. The height of the root is the length of the longest simple path to one of these $k$ leaves, which must have length $m$. It is clear from the way we defined $m$ that $m = \lfloor \lg n\rfloor$.


Show that in any subtree of a max-heap, the root of the subtree contains the largest value occuring anywhere in the subtree.

If the largest element in the subtree were somewhere other than the root, it has a parent that is in the subtree. So, it is larger than it's parent, so, the heap property is violated at the parent of the maximum element in the subtree.


Where in a max-heap might the smallest element reside, assuming that all elements are distinct?

In any of the leaves, that is, elements with index $\lfloor n / 2 \rfloor + k$, where $k \geq 1$ (see exercise 6.1-7), that is, in the second half of the heap array.


Is an array that is in sorted order a min-heap?

Yes. For any index $i$, both $\text{LEFT}(i)$ and $\text{RIGHT}(i)$ are larger and thus the elements indexed by them are greater or equal to $A[i]$ (because the array is sorted.)


Is the array with values $\langle 23, 17, 14, 6, 13, 10, 1, 5, 7, 12 \rangle$ a max-heap?

No. Since $\text{PARENT}(7)$ is $6$ in the array. This violates the max-heap property.


Show that, with the array representation for sorting an $n$-element heap, the leaves are the nodes indexed by $\lfloor n / 2 \rfloor + 1, \lfloor n / 2 \rfloor + 2, \ldots, n$.

Let's take the left child of the node indexed by $\lfloor n / 2 \rfloor + 1$.

$$ \begin{aligned} \text{LEFT}(\lfloor n / 2 \rfloor + 1) & = 2(\lfloor n / 2 \rfloor + 1) \\ & > 2(n / 2 - 1) + 2 \\ & = n - 2 + 2 \\ & = n. \end{aligned} $$

Since the index of the left child is larger than the number of elements in the heap, the node doesn't have childrens and thus is a leaf. Same goes for all nodes with larger indices.

Note that if we take element indexed by $\lfloor n / 2 \rfloor$, it will not be a leaf. In case of even number of nodes, it will have a left child with index $n$ and in the case of odd number of nodes, it will have a left child with index $n - 1$ and a right child with index $n$.

This makes the number of leaves in a heap of size $n$ equal to $\lceil n / 2 \rceil$.