Homework 5: Binary Search Trees and Heaps
Due: Sunday, November 11, 8:59:59pm
We are going retro with this homework. Fill in your answers in the
following plain text file. We are using ASCII art for binary trees, so
just use your favorite text editor to answer the questions (e.g., emacs,
vim, nano). Please DO NOT USE TABS,
it messes up the ASCII art. Also, Unix line endings please.
Please only enter text answers in the space provided, and for the tree
diagrams, only modify the 'X's: do not change any other part of the
diagram, nor actually edit out any branches (you are instructed
to replace the 'X's with blanks instead).
How to submit:
Here's the link for the file shown below: hw5.txt.
Edit the file and copy into the hw5 submission directory on GL
as usual.
<![CDATA[
Name:
Userid:
CMSC 341 Fall 2018 Homework 5
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Question #1:
-------------------------------------------------------------------
How unbalanced can a red-black tree be? That is, how much taller
can the left subtree be compared to the right subtree? Explain your
answer.
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Question #2:
-------------------------------------------------------------------
Using the ascii-art binary tree below, make a red-black tree with
height 4 (that's true tree height, not black height) using the
fewest number of nodes. That is, for each node, either replace the
"X" with a "b" for black, "R" for red or spaces to indicate the
absence of a node. (We want you to use lower case "b" for black
because "B" and "R" look too similar.) Note that there should be
at least one "b" or "R" at the lowest level. Please don't use
tabs: people have different tab stop settings and that would ruin
the ascii art.
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ / \
X X X X X X X X X X X X X X X X
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Question #3:
-------------------------------------------------------------------
In a max heap (largest value at the root), when we call deleteMax,
the root is replaced by the last item in the heap and "trickles down" to
the correct position.
In the binary heap below, the last item happens to be 1. Replace the X's
with actual positive values (no duplicates), so you have a legal max
heap and after each of the 4 calls to deleteMax, the number 1 trickles
down to the last position again.
"Draw" the resulting heap after each deleteMax. (You might want to copy
over your original heap four times instead of entering the numbers
again.)
Original Heap:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ / \
X X X X X X X X X X X X X X X 1
After first deleteMax:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ /
X X X X X X X X X X X X X X 1
After second deleteMax:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \
X X X X X X X X X X X X X 1
After third deleteMax:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ /
X X X X X X X X X X X X 1
After fourth deleteMax:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \
X X X X X X X X X X X 1
-------------------------------------------------------------------
Question #4:
-------------------------------------------------------------------
A binary heap can support a version of deletion in O(log n) time.
In this version, you are given a location in the heap to delete. So,
the delete instruction is given an index into the array and asked to
delete the item in that location. The item that is removed is replaced
by the last item in the heap. (We might as well do that since the
shape of the heap is completely determined by the number of items).
Of course the replacement item might violate the heap property ---
i.e., its key may be too small or too large. In this case we may
need to "bubble up" or "trickle down" to fix the heap.
In may not be obvious that both "bubbling up" and "trickling down" may
be needed. In the two binary heaps below, enter numbers in the nodes
for a MAX heap so that in the first case, after the key 30 is deleted,
the replacement item bubbles up and in the second case, after the key 30
is deleted, the replacement item trickles down.
Both heaps have to be MAX heaps (but they, of course, cannot be the same
heap). Also, show the resulting heap after deletion.
First max heap: deleting 30 makes the replacement item bubble up:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X 30 X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ / \
X X X X X X X X X X X X X X X X
Result of first heap, after deleting 30:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ /
X X X X X X X X X X X X X X X
Second max heap: deleting 30 makes the replacement item trickle down.
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X 30 X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ / \
X X X X X X X X X X X X X X X X
Result of second heap, after deleting 30:
X
/ \
___________/ \___________
/ \
/ \
X X
/ \ / \
___/ \___ ___/ \___
/ \ / \
/ \ / \
X X X X
_/ \_ _/ \_ _/ \_ _/ \_
/ \ / \ / \ / \
X X X X X X X X
/ \ / \ / \ / \ / \ / \ / \ /
X X X X X X X X X X X X X X X
]]>