(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
Related
This is task I found from old exam and I try it for solved because they can ask similar question for me on friday.
For solution I have cheap solution but I think it all matter of definition of binary search tree.
I make first tree:
1
\
1
\
1
and here is second tree
1
/
1
/
1
When you make pre order traversal you have same output for both tree.. because same element, and both have degenerated tree. But you don't have same tree! So statement is false.
Only problem is are my tree binary search tree... I think yes because binary search tree the element can have greater / lower equal element?
Please halp when I asked our teacher he say I can ask him when vacation is over but when vacation is over my exam is over.... No good thing for me.
Your answer is perfectly fine given the standard definition of BSTs. By the standard definition, BSTs can have repeated elements and identical elements can go in either subtree.
If the question intends the case where there are no duplicates, or where duplicates must exist in only the left (or only the right) subtree, then a pre-order traversal is sufficient to reconstruct the tree.
If duplicates are not allowed, construct the tree recursively as follows: make the root the first node, and then make the left and right subtrees recursively using as input traversals the portions of the original traversal with nodes less than (for left subtree) and greater than (for right subtree) the root node. If duplicates are allowed but are constrained to either the left or the right subtree, then use the same procedure but adding "or equal" to either the less than or the greater than, but not both.
I came upon two resources and they appear to say the basic definition in two ways.
Source 1 (and one of my professor) says:
All leaves are at the same level and all non-leaf nodes have two child nodes.
Source 2 (and 95% of internet) says:
A full binary tree (sometimes referred to as a proper or plane binary tree) is a tree in which every node in the tree has either 0 or 2 children.
Now following Source 2,
becomes a binary tree but not according to Source 1 as the leaves are not in the same level.
So typically they consider trees like,
as Full Binary Tree.
I may sound stupid but I'm confused what to believe. Any help is appreciated. Thanks in advance.
There are three main concepts: (1) Full binary tree (2) Complete binary tree and (3) Perfect binary tree. As you said, full binary tree is a tree in which all nodes have either degree 2 or 0. However, a complete binary tree is one in which all levels except possibly the last level are filled from left to right. Finally, a perfect binary tree is a full binary tree in which all leaves are at the same depth. For more see the wikipedia page
My intuition for the term complete here is that given a fixed number of nodes, a complete binary tree is made by completing each level from left to right except possibly the last one, as the number of nodes may not always be of the form 2^n - 1.
I think the issue is, what's the purpose of making the definition? Usually, the reason for defining full binary tree in the way that appears in Wikipedia is to be able to introduce and prove the Full Binary Tree Theorem:
The total number of nodes N in a full binary tree with I internal nodes is 2 I + 1.
(There are several equivalent formulations of this theorem in terms of the number of interior nodes, number of leaf nodes, and total number of nodes.) The proof of this theorem does not require that all the leaf nodes be at the same level.
What one of your professors is describing is something I would call a perfect binary tree, or, equivalently, a full, complete binary tree.
Suppose we have a set of binary trees with their inorder and preorder traversals given,and where no tree is a subtree of another tree in the given set. Now another binary tree Q is given.find whether it can be formed by joining the binary trees from the given set.(while joining each tree in the set should be considered atmost once) joining operation means:
Pick the root of any tree in the set and hook it to any vertex of another tree such that the resulting tree is also a binary tree.
Can we do this using LCA (least common ancestor)?or does it needs any special datastructure to solve?
I think a Binary tree structure should be enough. I don't think you NEED any other special data structure.
And I don't understand how you would use LCA for this. As far as my knowledge goes, LCA is used for knowing the lowest common Ancestor for two NODES in the same tree. It would not help in comparing two trees. (which is what I would do to check if Q can be formed)
My solution in words.
The tree Q that has to be checked if it can be made from the set of trees, So I would take a top-down approach. Basically comparing Q with the possible trees formed from the set.
Logic:
if Q.root does not match with any of the roots of the trees in the set (A,B,C....Z...), No solution possible.
if Q.root matches a Tree root (say A) check corresponding children and mark A as used/visited. (Which is what I understand from the question: a tree can be used only once)
We should continue with A in our solution only if all of Q's children match the corresponding children of A. (I would do Depth First traversal, Breadth First would work as well).
We can add append a new tree from the set (i.e. append a new root (tree B) only at leaf nodes of A as we have to maintain binary tree). Keep track of where the B was appended.
Repeat same check with corresponding children comparison as done for A. If no match, remove B and try to add C tree at the place where B was Added.
We continue to do this till we run out of nodes in Q. (unless we want IDENTICAL MATCH, in which case we would try other tree combinations other than the ones that we have, which match Q but have more nodes).
Apologies for the lengthy verbose answer. (I feel my pseudo code would be difficult to write and be riddled with comments to explain).
Hope this helps.
An alternate solution: Will be much less efficient (try only if there are relatively less number of trees) : forming all possible set of trees ( first in 2s then 3s ....N) and and Checking the formed trees if they are identical to Q.
the comparing part can be referred here:
http://www.geeksforgeeks.org/write-c-code-to-determine-if-two-trees-are-identical/
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.