I saw an answer here with the idea implemented in Python (not very familiar with Python) - I was looking for a more general algorithm.
EDIT:
For clarification:
Say we are given a list of integer keys: 23 44 88 12 74 32 7 39 10
That list was chosen arbitrarily. We are to create an almost complete (or complete) binary search tree from that list. There is supposed to be only one such tree...how do we find it?
A binary search tree is constructed so that all items on a node's left subtree are less than the node, and all nodes on the right subtree are greater than the node.
A complete (or almost complete) binary tree is one in which all levels except possibly the last are completely full, and the bottom level is filled to the left.
So, for example, this is an almost-complete binary search tree:
4
/ \
2 5
/ \
1 3
This is not:
3
/ \
2 4
/ \
1 5
Because the bottom level of the tree is not filled from the left.
If the number of items is one less than a power of two (i.e. 3, 7, 15, etc.), then building the tree is easy. Start by sorting the list. Then, take the middle element as the root. So if you have [1,2,3,4,5,6,7], and the root node is 4.
You do the same thing recursively for the right and left halves of the array.
If the number of items is not one less than a power of two, you have to adjust the starting point (the root node) so that the bottom row is left-filled. Note that you might have to apply that adjustment recursively, as well, whenever your subtree length is not one less than a power of two.
Since this is a homework assignment, I'll leave that for you to figure out.
My goal is to create a BST (Binary Search Tree) for compiler of pseudo-pascal code.
From what I know BST works like this: I have a root, which has some key - in dependency on MY GENERATED KEY (if it is greater or lesser than root key) I insert a node to the left or right from the root.
Obviously, if I use numbers for keys, the tree will go like:
1
\
2
\
3
What I want is a balanced BST, something like this:
1
/ \
2 3
... etc.
Which is obviously not gonna happen with those keys. How should keys for nodes look like when I want to achieve this structure? Is there a way to do this without remembering the last inserted key?
That also brings me to the question what "key" should be used for absolute root of the tree? So I can add keys to the left and to the right from it?
When you insert a node, you travel from root node down to find proper insertion point:
1 1 1
\ \
2 2
\
3
After insertion, when you return back to root, on the way you check that each left and right nodes are balanced enough. If not, you rotate them:
1 2
\ / \
2 1 3
\
3
This means that first node of deeper branch comes the new root node, and old root node will become child of this node.
So root node will change when necessary, and will always have the "most middle" value.
Apply the concept of AVL tree to balance the tree structure.
According to your requirement:
1
/ \
2 3
The above tree is constructed just considering a binary tree not a binary Search Tree.
However above mentioned tree can be constructed by top-to-down and from left-to-right fashion.
I am preparing for an exam and I have stumbled on the following question:
Draw the binary search tree that would result if data were to be added in the following order:
10,9,8,7,6,5,4,3
Why is the tree that results unsuitable for efficient searching?
My Answer:
I would have thought when creating a BST that we start with the value 10 as the root node then add 9 as the left sub tree value on the first level. Then 8 to the left subtree of 9 and so on. I don't know why this makes it inefficient for searching though. Any ideas?
Since the values are in decreasing order, they get added to the left at each level, practically leaving you with a linked list, which takes O(N) to search, instead of the preferred O(logN) of a BST.
Drawing:
10
/
9
/
8
/
7
/
6
/
5
/
4
/
3
This would create a linked-list, since it would just be a series of nodes; which is a heavily unbalanced tree.
You should look up red-black trees. They have the same time-complexities, but it will constantly move around nodes, so that it is always forming a triangular shape. This will keep the tree balanced.
This is inefficient because the node will always be added to the left subtree of the prior node. Making a search check every every node in the list until it finds the result even though the answer will always be to the left thus actually making it take more computations than just simply having a list that is searched through a loop.
I'm working through a past exam paper for my advanced programming course and I've gotten stuck at this question
What property must the values in a binary search tree satisfy? How many different binary search trees are there containing the three values 1 2 3? Explain your answer.
I can answer the first part easily enough but the second bit, about the number of possible trees has me stumped. My first instinct is to say that there is only a single tree possible, with 2 as the root because the definition says so, but this question is work 8 marks out of a total of 100 for the entire paper, so I can only assume that it's a trick question, and there's a more subtle explanation, but there's nothing in the lecture notes that explains this. Does anyone know who to answer this question?
The question doesn't say that the tree is balanced, so think about whether 1 or 3 can be at the root node.
Try to think about all possible binary trees with these three nodes. How many of those trees fulfill the property of binary search tree?
I think that a trick is that a tree can be a degenerate one (effectively, a linked list of elements):
1
\
2
\
3
And variations thereof.
Also, are these trees considered to be identical ?
2 2
/ \ / \
3 1 1 3
If I remember correctly, the root of the tree does not have to be the "middle element". Thus there are a few more combinations of trees:
2
1 3
or
1
2
3
or
1
3
2
or
3
2
1
or
3
1
2
Maybe I forget a few, but I think you'll get the idea. Just for my notation: Newline meets get down in the tree, right and left of the upperline showes whether it is right or left of its parent node ;)
I'm trying to find the definition of a binary search tree and I keep finding different definitions everywhere.
Some say that for any given subtree the left child key is less than or equal to the root.
Some say that for any given subtree the right child key is greater than or equal to the root.
And my old college data structures book says "every element has a key and no two elements have the same key."
Is there a universal definition of a bst? Particularly in regards to what to do with trees with multiple instances of the same key.
EDIT: Maybe I was unclear, the definitions I'm seeing are
1) left <= root < right
2) left < root <= right
3) left < root < right, such that no duplicate keys exist.
Many algorithms will specify that duplicates are excluded. For example, the example algorithms in the MIT Algorithms book usually present examples without duplicates. It is fairly trivial to implement duplicates (either as a list at the node, or in one particular direction.)
Most (that I've seen) specify left children as <= and right children as >. Practically speaking, a BST which allows either of the right or left children to be equal to the root node, will require extra computational steps to finish a search where duplicate nodes are allowed.
It is best to utilize a list at the node to store duplicates, as inserting an '=' value to one side of a node requires rewriting the tree on that side to place the node as the child, or the node is placed as a grand-child, at some point below, which eliminates some of the search efficiency.
You have to remember, most of the classroom examples are simplified to portray and deliver the concept. They aren't worth squat in many real-world situations. But the statement, "every element has a key and no two elements have the same key", is not violated by the use of a list at the element node.
So go with what your data structures book said!
Edit:
Universal Definition of a Binary Search Tree involves storing and search for a key based on traversing a data structure in one of two directions. In the pragmatic sense, that means if the value is <>, you traverse the data structure in one of two 'directions'. So, in that sense, duplicate values don't make any sense at all.
This is different from BSP, or binary search partition, but not all that different. The algorithm to search has one of two directions for 'travel', or it is done (successfully or not.) So I apologize that my original answer didn't address the concept of a 'universal definition', as duplicates are really a distinct topic (something you deal with after a successful search, not as part of the binary search.)
If your binary search tree is a red black tree, or you intend to any kind of "tree rotation" operations, duplicate nodes will cause problems. Imagine your tree rule is this:
left < root <= right
Now imagine a simple tree whose root is 5, left child is nil, and right child is 5. If you do a left rotation on the root you end up with a 5 in the left child and a 5 in the root with the right child being nil. Now something in the left tree is equal to the root, but your rule above assumed left < root.
I spent hours trying to figure out why my red/black trees would occasionally traverse out of order, the problem was what I described above. Hopefully somebody reads this and saves themselves hours of debugging in the future!
All three definitions are acceptable and correct. They define different variations of a BST.
Your college data structure's book failed to clarify that its definition was not the only possible.
Certainly, allowing duplicates adds complexity. If you use the definition "left <= root < right" and you have a tree like:
3
/ \
2 4
then adding a "3" duplicate key to this tree will result in:
3
/ \
2 4
\
3
Note that the duplicates are not in contiguous levels.
This is a big issue when allowing duplicates in a BST representation as the one above: duplicates may be separated by any number of levels, so checking for duplicate's existence is not that simple as just checking for immediate childs of a node.
An option to avoid this issue is to not represent duplicates structurally (as separate nodes) but instead use a counter that counts the number of occurrences of the key. The previous example would then have a tree like:
3(1)
/ \
2(1) 4(1)
and after insertion of the duplicate "3" key it will become:
3(2)
/ \
2(1) 4(1)
This simplifies lookup, removal and insertion operations, at the expense of some extra bytes and counter operations.
In a BST, all values descending on the left side of a node are less than (or equal to, see later) the node itself. Similarly, all values descending on the right side of a node are greater than (or equal to) that node value(a).
Some BSTs may choose to allow duplicate values, hence the "or equal to" qualifiers above. The following example may clarify:
14
/ \
13 22
/ / \
1 16 29
/ \
28 29
This shows a BST that allows duplicates(b) - you can see that to find a value, you start at the root node and go down the left or right subtree depending on whether your search value is less than or greater than the node value.
This can be done recursively with something like:
def hasVal (node, srchval):
if node == NULL:
return false
if node.val == srchval:
return true
if node.val > srchval:
return hasVal (node.left, srchval)
return hasVal (node.right, srchval)
and calling it with:
foundIt = hasVal (rootNode, valToLookFor)
Duplicates add a little complexity since you may need to keep searching once you've found your value, for other nodes of the same value. Obviously that doesn't matter for hasVal since it doesn't matter how many there are, just whether at least one exists. It will however matter for things like countVal, since it needs to know how many there are.
(a) You could actually sort them in the opposite direction should you so wish provided you adjust how you search for a specific key. A BST need only maintain some sorted order, whether that's ascending or descending (or even some weird multi-layer-sort method like all odd numbers ascending, then all even numbers descending) is not relevant.
(b) Interestingly, if your sorting key uses the entire value stored at a node (so that nodes containing the same key have no other extra information to distinguish them), there can be performance gains from adding a count to each node, rather than allowing duplicate nodes.
The main benefit is that adding or removing a duplicate will simply modify the count rather than inserting or deleting a new node (an action that may require re-balancing the tree).
So, to add an item, you first check if it already exists. If so, just increment the count and exit. If not, you need to insert a new node with a count of one then rebalance.
To remove an item, you find it then decrement the count - only if the resultant count is zero do you then remove the actual node from the tree and rebalance.
Searches are also quicker given there are fewer nodes but that may not be a large impact.
For example, the following two trees (non-counting on the left, and counting on the right) would be equivalent (in the counting tree, i.c means c copies of item i):
__14__ ___22.2___
/ \ / \
14 22 7.1 29.1
/ \ / \ / \ / \
1 14 22 29 1.1 14.3 28.1 30.1
\ / \
7 28 30
Removing the leaf-node 22 from the left tree would involve rebalancing (since it now has a height differential of two) the resulting 22-29-28-30 subtree such as below (this is one option, there are others that also satisfy the "height differential must be zero or one" rule):
\ \
22 29
\ / \
29 --> 28 30
/ \ /
28 30 22
Doing the same operation on the right tree is a simple modification of the root node from 22.2 to 22.1 (with no rebalancing required).
In the book "Introduction to algorithms", third edition, by Cormen, Leiserson, Rivest and Stein, a binary search tree (BST) is explicitly defined as allowing duplicates. This can be seen in figure 12.1 and the following (page 287):
"The keys in a binary search tree are always stored in such a way as to satisfy the binary-search-tree property: Let x be a node in a binary search tree. If y is a node in the left subtree of x, then y:key <= x:key. If y is a node in the right subtree of x, then y:key >= x:key."
In addition, a red-black tree is then defined on page 308 as:
"A red-black tree is a binary search tree with one extra bit of storage per node: its color"
Therefore, red-black trees defined in this book support duplicates.
Any definition is valid. As long as you are consistent in your implementation (always put equal nodes to the right, always put them to the left, or never allow them) then you're fine. I think it is most common to not allow them, but it is still a BST if they are allowed and place either left or right.
I just want to add some more information to what #Robert Paulson answered.
Let's assume that node contains key & data. So nodes with the same key might contain different data.
(So the search must find all nodes with the same key)
left <= cur < right
left < cur <= right
left <= cur <= right
left < cur < right && cur contain sibling nodes with the same key.
left < cur < right, such that no duplicate keys exist.
1 & 2. works fine if the tree does not have any rotation-related functions to prevent skewness.
But this form doesn't work with AVL tree or Red-Black tree, because rotation will break the principal.
And even if search() finds the node with the key, it must traverse down to the leaf node for the nodes with duplicate key.
Making time complexity for search = theta(logN)
3. will work well with any form of BST with rotation-related functions.
But the search will take O(n), ruining the purpose of using BST.
Say we have the tree as below, with 3) principal.
12
/ \
10 20
/ \ /
9 11 12
/ \
10 12
If we do search(12) on this tree, even tho we found 12 at the root, we must keep search both left & right child to seek for the duplicate key.
This takes O(n) time as I've told.
4. is my personal favorite. Let's say sibling means the node with the same key.
We can change above tree into below.
12 - 12 - 12
/ \
10 - 10 20
/ \
9 11
Now any search will take O(logN) because we don't have to traverse children for the duplicate key.
And this principal also works well with AVL or RB tree.
Working on a red-black tree implementation I was getting problems validating the tree with multiple keys until I realized that with the red-black insert rotation, you have to loosen the constraint to
left <= root <= right
Since none of the documentation I was looking at allowed for duplicate keys and I didn't want to rewrite the rotation methods to account for it, I just decided to modify my nodes to allow for multiple values within the node, and no duplicate keys in the tree.
Those three things you said are all true.
Keys are unique
To the left are keys less than this one
To the right are keys greater than this one
I suppose you could reverse your tree and put the smaller keys on the right, but really the "left" and "right" concept is just that: a visual concept to help us think about a data structure which doesn't really have a left or right, so it doesn't really matter.
1.) left <= root < right
2.) left < root <= right
3.) left < root < right, such that no duplicate keys exist.
I might have to go and dig out my algorithm books, but off the top of my head (3) is the canonical form.
(1) or (2) only come about when you start to allow duplicates nodes and you put duplicate nodes in the tree itself (rather than the node containing a list).
Duplicate Keys
• What happens if there's more than one data item with
the same key?
– This presents a slight problem in red-black trees.
– It's important that nodes with the same key are distributed on
both sides of other nodes with the same key.
– That is, if keys arrive in the order 50, 50, 50,
• you want the second 50 to go to the right of the first one, and the
third 50 to go to the left of the first one.
• Otherwise, the tree becomes unbalanced.
• This could be handled by some kind of randomizing
process in the insertion algorithm.
– However, the search process then becomes more complicated if
all items with the same key must be found.
• It's simpler to outlaw items with the same key.
– In this discussion we'll assume duplicates aren't allowed
One can create a linked list for each node of the tree that contains duplicate keys and store data in the list.
The elements ordering relation <= is a total order so the relation must be reflexive but commonly a binary search tree (aka BST) is a tree without duplicates.
Otherwise if there are duplicates you need run twice or more the same function of deletion!