This exercise is from the book "algorithms in c++". Suppose that there is a free tree like this
First Question: Could we represent this free tree as a binary tree?
I think that we cannot represent this free tree as a binary tree
because there are some nodes that have more than 2 children.
Second Question: How many ordered trees could we represent from this free tree?
I cannot understand the question. There aren't keys in the nodes in order to decide how to create an ordered tree.
If I understood the question correctly, the task is that of representing a tree as a binary tree, i.e. to use a data structure for binary trees to represent arbitrary trees. Speaking more structurally, the question demands for a method to injectively map arbitrary trees to binary trees.
The technique is described here; the basic idea is to represent the children c_1,...c_nof a node a, which may be more than two, by an arbitrarily chosen child, say c_1, which becomes the left child of a. As the right child of c_1, the next child, c_2 is stored; and so on. This means that for each node, one child is stored in the left subtree, while in the right subtree by chosing always the right child, the 'siblings' of 'alternatives' of the child are stored. The approach can be sketched as follows. Please bear with me for the relatively poor textual representation.
arbitrary tree
a
/ | | \
c_1 c_2 c_3 c-4
binary tree
a
/
c_1
\
c_2
\
c_3
\
c_4
Related
Just been learning about binary trees in school, and two rules of binary trees is that
every node has at most 2 child nodes
there exists linear ordering defined for the children of each node (ordered pair)
Now, all types of binary trees (full, complete, etc.) are binary trees so they must satisfy these 2 conditions.
However, I saw on GeeksForGeeks this example:
How is 'linear ordering', ordered pair, defined here?
It seems that for the sibling nodes in this picture, some left ones are larger than the right one, some right nodes are larger than the left one.
If asked to check if a given tree is a binary tree, how do I make sure of the second property, that the children of each node have to be ordered?
Thanks
This is one of the complicated ways to introduce a binary tree.
two rules of binary trees is that
every node has at most 2 child nodes
there exists linear ordering defined for the children of each node (ordered pair)
Simple ways of introducing binary trees I could think of are "at most two children and no cycles" or "at most two children and unique path between any pair of vertices".
But fine. You bring up the point of linear order. Lets discuss that.
Here
A linear ordering on a finite collection of objects may be described
as follows: each object has exactly one immediate predecessor object
and one immediate successor object with two exceptions: A first object
has no predecessor and a last object has no successor.
If you have learnt about traversal so far, with the above definition, I would take binary tree traversals as linear order - preorder, postorder, inorder, level order. This applies to all types of binary trees (full, complete, etc.) which includes the complete binary tree you posted as an image.
(From: Data Structures Using C By Aaron M. Tenenbaum):
"A complete binary tree of depth d is the strictly binary tree all of whose leaves are at level d."
So, by that meaning the following tree should not be complete binary tree, right?
http://cs-study.blogspot.de/2012/11/complete-binary-tree.html
But, it is according to wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes at the last level h.
Please clarify my confusion.
My interpretation:
On level 1..d-1 there are only nodes, and all of them must exist.
At level d, only leaves exists, and they must be filled from left to right
Nodes without children are not considered leaves on level d-1
Given a set of values, it's possible for there to be many different possible binary search trees that can be formed from those values. For example, for the values 1, 2, and 3, there are five BSTs we can make from those values:
1 1 2 3 3
\ \ / \ / /
2 3 1 3 1 2
\ / \ /
3 2 2 1
Many data structures that are based on balanced binary search trees use tree rotations as a primitive for reshaping a BST without breaking the required binary search tree invariants. Tree rotations can be used to pull a node up above its parent, as shown here:
rotate
u right v
/ \ -----> / \
v C A u
/ \ <----- / \
A B rotate B C
left
Given a BST containing a set of values, is it always possible to convert that BST into any arbitrary other BST for the same set of values? For example, could we convert between any of the five BSTs above into any of the other BSTs just by using tree rotations?
The answer to your question depends on whether you are allowed to have equal values in the BST that can appear different from one another. For example, if your BST stores key/value pairs, then it is not always possible to turn one BST for those key/value pairs into a different BST for the same key/value pairs.
The reason for this is that the inorder traversal of the nodes in a BST remains the same regardless of how many tree rotations are performed. As a result, it's not possible to convert from one BST to another if the inorder traversal of the nodes would come out differently. As a very simple case, suppose you have a BST holding two copies of the number 1, each of which is annotated with a different value (say, A or B). In that case, there is no way to turn these two trees into one another using tree rotations:
1:a 1:b
\ \
1:b 1:a
You can check this by brute-forcing the (very small!) set of possible trees you can make with the rotations. However, it suffices to note that an inorder traversal of the first tree gives 1:a, 1:b and an inorder traversal of the second tree gives 1:b, 1:a. Consequently, no number of rotations will suffice to convert between the trees.
On the other hand, if all the values are different, then it is always possible to convert between two BSTs by applying the right number of tree rotations. I'll prove this using an inductive argument on the number of nodes.
As a simple base case, if there are no nodes in the tree, there is only one possible BST holding those nodes: the empty tree. Therefore, it's always possible to convert between two trees with zero nodes in them, since the start and end tree must always be the same.
For the inductive step, let's assume that for any two BSTs of 0, 1, 2, .., n nodes with the same values, that it's always possible to convert from one BST to another using rotations. We'll prove that given any two BSTs made from the same n + 1 values, it's always possible to convert the first tree to the second.
To do this, we'll start off by making a key observation. Given any node in a BST, it is always possible to apply tree rotations to pull that node up to the root of the tree. To do this, we can apply this algorithm:
while (target node is not the root) {
if (node is a left child) {
apply a right rotation to the node and its parent;
} else {
apply a left rotation to the node and its parent;
}
}
The reason that this works is that every time a node is rotated with its parent, its height increases by one. As a result, after applying sufficiently many rotations of the above forms, we can get the root up to the top of the tree.
This now gives us a very straightforward recursive algorithm we can use to reshape any one BST into another BST using rotations. The idea is as follows. First, look at the root node of the second tree. Find that node in the first tree (this is pretty easy, since it's a BST!), then use the above algorithm to pull it up to the root of the tree. At this point, we have turned the first tree into a tree with the following properties:
The first tree's root node is the root node of the second tree.
The first tree's right subtree contains the same nodes as the second tree's right subtree, but possibly with a different shape.
The first tree's left subtree contains the same nodes as the second tree's left subtree, but possibly with a different shape.
Consequently, we could then recursively apply this same algorithm to make the left subtree have the same shape as the left subtree of the second tree and to make the right subtree have the same shape as the right subtree of the second tree. Since these left and right subtrees must have strictly no more than n nodes each, by our inductive hypothesis we know that it's always possible to do this, and so the algorithm will work as intended.
To summarize, the algorithm works as follows:
If the two trees are empty, we are done.
Find the root node of the second tree in the first tree.
Apply rotations to bring that node up to the root.
Recursively reshape the left subtree of the first tree to have the same shape as the left subtree of the second tree.
Recursively reshape the right subtree of the first tree to have the same shape as the right subtree of the second tree.
To analyze the runtime of this algorithm, note that applying steps 1 - 3 requires at most O(h) steps, where h is the height of the first tree. Every node will be brought up to the root of some subtree exactly once, so we do this a total of O(n) times. Since the height of an n-node tree is never greater than O(n), this means that the algorithm takes at most O(n2) time to complete. It's possible that it will do a lot better (for example, if the two trees already have the same shape, then this runs in time O(n)), but this gives a nice worst-case bound.
Hope this helps!
For binary search trees this can actually be done in O(n).
Any tree can be "straightened out", ie put into a form in which all nodes are either the root or a left child.
This form is unique (reading down from root gives the ordering of the elements)
A tree is straightened out as follows:
For any right child, perform a left rotation about itself. This decreases the number of right children by 1, so the tree is straightened out in O(n) rotations.
If A can be straightened out into S in O(n) rotations, and B into S in O(n) rotations, then since rotations are reversible one can turn A -> S -> B in O(n) rotations.
I see this definition of a binary tree in Wikipedia:
Another way of defining binary trees is a recursive definition on directed graphs. A binary tree is either:
A single vertex.
A graph formed by taking two binary trees, adding a vertex, and adding an edge directed from the new vertex to the root of each binary tree.
How then is it possible to have a binary tree with one root and one left son, like this:
O
/
O
This is a binary tree, right? What am I missing here?
And please don't just say "Wikipedia can be wrong", I've seen this definition in a few other places as well.
Correct. A tree can be empty (nil)
Let's assume you have two trees: one, that has one vertex, and one which is empty (nil). They look like this:
O .
Notice that I used a dot for the (nil) tree.
Then I add a new vertex, and edges from the new vertex to the existing two trees (notice that we do not take edges from the existing trees and connect them to the new vertes - it would be impossible.). So it looks like it now:
O
/ \
O .
Since edges leading to (nil) are not drawn, here it is what is at the end:
O
/
O
I hope it clarifies.
It depends on the algorithm you use for binary-tree: as for icecream, there are many flavors :)
One example is when you have a mix of node pointers and leaf pointers on a node, and a balancing system that decide to create a second node (wether it's the root or the other) when you are inserting new values on a full node: instead of creating a root and 2 leafs, by splitting it, it's much more economical to create just another node.
Wikipedia can be wrong. Binary trees are finite data structures, a subtree must be allowed to be empty otherwise binary trees would be infinite. The base case for the recursive definition of a binary tree must allow either a single node or the empty tree.
Section 14.4 of Touch of Class: An Introduction to Programming Well Using Objects
and Contracts, by Bertrand Meyer, Springer Verlag, 2009. © Bertrand Meyer, 2009. has a better recursive definition of a binary tree
Definition: binary tree
A binary tree over G, for an arbitrary data type G, is a finite set of items called
nodes, each containing a value of type G, such that the nodes, if any, are
divided into three disjoint parts:
• A single node, called the root of the binary tree.
• (Recursively) two binary trees over G, called the left subtree and right subtree.
The definition explicitly allows a binary tree to be empty (“the nodes, if any”).
Without this, of course, the recursive definition would lead to an infinite
structure, whereas our binary trees are, as the definition also prescribes, finite.
If not empty, a binary tree always has a root, and may have: no subtree; a
left subtree only; a right subtree only; or both.
I have some questions on binary trees:
Wikipedia states that a binary tree is complete when "A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible." What does the last "as far left as possible" passage mean?
A well-formed binary tree is said to be "height-balanced" if (1) it is empty, or (2) its left and right children are height-balanced and the height of the left tree is within 1 of the height of the right tree, taken from How to determine if binary tree is balanced?, is this correct or there's "jitter" on the 1-value? I read on the answer I linked that there could be also a difference factor of 4 between the height of the right and the left tree
Do the complete and height-balanced definitions just apply to binary tree or just any other tree?
Following the reference of the definition in wikipedia, I got to
this page. The definition was taken from there but modified:
Definition: A binary tree in which every level, except possibly the deepest, is completely filled. At depth n, the height of the
tree, all nodes must be as far left as possible.
It continues with a note below though,
A complete binary tree has 2k nodes at every depth k < n and between 2n and 2^(n+1) - 1 nodes altogether.
Sometimes, definitions vary according to convenience (be useful for something). That passage might be a variation which, as I understand, requires leaf nodes to fill first the left side of the deepest level (that is, fill from left to right). The definition that I usually found is exactly as described above but without that
passage.
Usually the definition taken for height-balanced tree is the one you
described. In other words:
A tree is balanced if and only if for every node the heights of its two subtrees differ by at most 1.
That definition was taken from here. Again, sometimes definitions are made more flexible to serve specific purposes. For example, the definition of an AVL tree says that
In an AVL tree, the heights of the two child subtrees of any node
differ by at most one
Still, I remember once I had to rewrite an algorithm so that the tree
would be considered height-balanced if the two child subtrees of any
node differed by at most 2. Note that the definition you gave is recursive, this is very common for binary trees.
In a tree whose number of children is variable, you wouldn't be able to say that it is complete (any parent could have the number of children that you want). Still, it can apply to n-ary trees (with a fixed amount of n children).
Do the complete and height-balanced definitions just apply to binary
tree or just any other tree?
Short answer: Yes, it can be extended to any n-ary tree.