I want to solve the problem given in the image below.
A New Algorithm to Represent a Given k-ary Tree into Its Equivalent Binary Tree.
Refer This Paper
In Simple words:
1. Create L to R sibling pointers at each level
2. Remove all but the leftmost child pointer of each node
3. Make the sibling pointer the right pointer.
Related
Can a Binary Tree / Binary Search Tree only have the parent and one of the left or right nodes?
Or is it mandatory to have both the left and right nodes?
Binary implies two. Binary trees have two children pointers, not necessarily with references to anything. Parent references are not necessary - that is up to implementation.
If nodes were mandatory to have both children elements, then the tree would be infinite and leaves would not exist.
So, options for a node are
leaf
left/right child only
both children
Of course, there are alternative ways to think about a "tree". For example, take the binary heap array implementation.
It is not mandatory, you can create other types of links if you want. Define up and down pointers like this.
For a node x define up pointer u(x) as:
if x is the right child, then u(x) is the parent of x
if x is left child or root, then u(x) is the lowest node on the right path of x
For a node x define down pointer d(x) as:
if x's left subtree does not exist, then d(x) = x
if left subtree exists, then d(x) points to the rightmost node of x's left subtree
This is so called ring representation of a tree.
Actually, there is nothing mandatory other than just one rule (i.e. the maximum number of children that any binary tree node can have is two).
Technically:
Even a NULL is considered as a tree node when we define TreeNode *root = NULL.
Though, it has no practical significance, we do pass them as argument while calling insert, delete and display methods.
So, as a coder, assuming that a binary tree will always comprise of at least a parent node can prove fatal if the sanity of root is not checked before processing a binary tree.
A binary search tree is also a binary tree with additional properties. Any Binary tree node can have either 0 or 1 or 2 child nodes. So to answer your question, its okay to have only left or right nodes
I'm implementing the Bentley-Ottmann algorithm
to find the set of segment intersection points,
unfortunately I didn't understand some things.
For example :
how can I get the neighbours of the segment Sj in the image.
I'm using a balanced binary search tree for the sweepLine status, but we store the segments in the leaves, after reading this wikipedia article I didn't find an explanation for this operation.
From the reference book (de Berg & al.: "Computational Geometry", ill. at p.25):
Suppose we search in T for the segment immediately to the left of some point p that lies on the sweep line.
At each internal node v we test whether p lies left or right of the segment stored at v.
Depending on the outcome we descend to the left or right subtree of v,
eventually ending up in a leaf.
Either this leaf, or the leaf immediately to the left of it, stores the segment we are searching for.
For my example if I follow this I will arrive at the leaf Sj but I will know just the leaf to the left i.e. Sk, how can I get Si?
Edit
I found this discussion that looks like my problem, unfortunately there are no answers about how can I implement some operations in such data structure.
The operation are:
inserting a node in such data structure.
deleting a node.
swapping two nodes.
searching for neighbours' node.
I know how to implement these operations in a balanced binary search tree when we store data too in internal node, but with this type of AVL I don't know if it is the same thing.
Thank you
I stumbled upon the same problem when reading Computational Geometry from DeBerg (see p. 25 for the quote and the image). My understanding is the following:
say you need the right neighbor of a segment S which is in the tree. If you store data in the nodes, the pseudo code is:
locate node S
if S has a right subtree:
return the left-most node of the right subtree of S
else if S is in the left sub-tree of any ancestor:
return the lowest/nearest such ancestor
else
return not found
If you store the data in the leaves, the pseudo-code becomes:
let p the point of S currently on the sweep line
let n the segment at the root of the tree
while n != null && n is not a leaf:
if n = S:
n = right child of S
else:
determine if p is on the right or left of n
update n accordingly (normal descent)
In the end, either n is null and it means there is no right neighbor, or n points to the proper leaf.
The same logic applies for the left neighbor.
Same as you, I have met the same problem while reading the de Berg & al.: "Computational Geometry". But I think The C++ Standard Template Library (STL) have an implantation called "map" which can do the job.
You just need to define some personalized class for line segment and event points and their comparison functions. Then, use std::map to build the tree and access the neighboring element using map.find() to get and iterator, and use iterator to gain access to the two neighbor element.
In the huffman coding algorithm, there's a lemma that says:
The binary tree corresponding to an optimal binary prefix code is full
But I can't figure out why. How can you prove this lemma?
Any binary code for data can be represented as a binary tree. The code is represented by the path from the root to the leaf, with a left edge representing a 0 in the prefix and a right one representing 1 (or vice versa.)
Keep in mind that for each symbol there is one leaf node.
To prove that an optimal code will be represented by a full binary tree, let's recall what a full binary tree is,
It is a tree where each node is either a leaf or has two chilren.
Let's assume that a certain code is optimal and is represented by a non-full tree.
So there is a certain vertex u with only a single child v. The edge between u and v adds the bit x to the prefix code of the symbols (at the leaves) in the subtree rooted at v.
From this tree I can remove the edge x and replace u with v, thus decreasing the length of the prefix code of all symbols in the subtree rooted at v by one. This reduces the number of bits in the representation of at least one symbol (when v is a singleton node.)
This shows that the tree didnt actually represent an optimal code, and my premise was wrong. Thus proving the lemma.
From wikipedia,
A full binary tree (sometimes 2-tree or strictly binary tree) is a tree in which every node other than the leaves has two children.
The way in which the tree for huffman code is produced will definitely produce a full binary tree. Because at each step of the algorithm, we remove the two nodes of highest priority (lowest probability) from the queue and create a new internal node with these two nodes as children.
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.
How to permutate all possible tree structure for fixed number of leaf for binary tree?
which algorithm to build all possible tree structure?
What is the number of all possible tree structure?
for binary expression tree, alphabet must be leaf
for example (a+b)*c
*
+ c
a b
the number of leaf is 3
we want to permutate to other structure such as
*
a +
b c
I guess to fix the order, the same order because i will permutate the operator node in later process. Just expect the structure. the number of operator node is dynamic
if you want to challenge more, you can use dynamic order
i also want to know dynamic order algorithm
now try to fix the order first
one more constraint i notice is that every internal node should have left and right leaf
My idea is recursively swap left and right node at each level except the leaf and save at each swap
I do not know whether is correct but i feel no missing