I am looking for a way to find lowest common ancestor in a binary tree.
Note : It is binary tree not binary search tree and we don't have any parent pointer. Moreover if we are trying to find a LCA for a parent(P) and any of its child(C) then the LCA will be immediate parent of P.
Related
I tried finding the difference between m way tree and the m way search tree. Most resources only tells about m way search tree and end up being on B tree or B+ trees.
My doubts are:-
Is it analogous to the binary tree and binary search tree?
I read somewhere that m way trees don't have any particular order and
every node has to be filled fully before moving to the new node.(complete
tree)
Is it analogous to the binary tree and binary search tree?
Yes
m way trees don't have any particular order
This is true
and every node has to be filled fully before moving to the new node.(complete tree)
Something like this describes a step in an algorithm, and has little to do with the data structure itself: nothing is "moving" in a data structure.
Definitions
In short: an m-way tree puts no conditions on the values stored in the nodes, while an m-way search tree does.
Reva Freedman, associate professor at Northern Illinois University has notes on Multiway Trees where four terms are defined in succession, each time indicating which additional requirements apply for the next term:
multi way tree,
m-way tree
m-way search tree
B-tree of order m
Multiway Trees
A multiway tree is a tree that can have more than two children. A
multiway tree of order m (or an m-way tree) is one in which a tree can
have m children.
As with the other trees that have been studied, the nodes in an m-way
tree will be made up of key fields, in this case m-1 key fields, and
pointers to children.
To make the processing of m-way trees easier, some type of order will
be imposed on the keys within each node, resulting in a multiway
search tree of order m ( or an m-way search tree). By definition an
m-way search tree is a m-way tree in which:
Each node has m children and m-1 key fields
The keys in each node are in ascending order.
The keys in the first i children are smaller than the ith key
The keys in the last m-i children are larger than the ith key
M-way search trees give the same advantages to m-way trees that binary
search trees gave to binary trees - they provide fast information
retrieval and update. However, they also have the same problems that
binary search trees had - they can become unbalanced, which means that
the construction of the tree becomes of vital importance.
B-Trees
An extension of a multiway search tree of order m is a B-tree of
order m. This type of tree will be used when the data to be
accessed/stored is located on secondary storage devices because they
allow for large amounts of data to be stored in a node.
A B-tree of order m is a multiway search tree in which:
The root has at least two subtrees unless it is the only node in the tree.
Each nonroot and each nonleaf node have at most m nonempty children and at least m/2 nonempty children.
The number of keys in each nonroot and each nonleaf node is one less than the number of its nonempty children.
All leaves are on the same level.
I am a beginner in the field of data structures, I am studying binary trees and in my textbook there's a tree which is not a binary tree but I am not able to make out why the tree is not a binary tree because every node in the tree has atmost two children.
According to Wikipedia definition of binary tree is "In computer science, a binary tree is a treedata structure in which each node has at most two children, which are referred to as the left child and the right child."
The tree in the picture seems to satisfy the condition as mentioned in the definition of binary tree.
I want an explanation for why the tree is not a binary tree?
This is not even a tree, let alone binary tree. Node I has two parents which violates the tree property.
I got the answer, This not even a tree because a tree is connected acyclic graph also a binary tree is a finite set of elements that is either empty or is partitioned into three disjoint subsets. The first subset contains a single element called the root of the tree. The other two subsets are themselves binary trees called the left and right subtrees of the original tree.
Here the word disjoint answers the problem.
It's not a binary tree because of node I
This can be ABEI or ACFI
This would mean the node can be represented by 2 binary numbers which is incorrect
Each node has either 0 or 1 parents. 0 in the case of the root node. 1 otherwise. I has 2 parents E and F
Strictly speaking, binary tree pre-order traversal is the same as binary tree depth first search. Binary tree in-order and post-order traversals are not the same as binary tree depth first search. Am I right?
All three tree traversals are produced by a depth first search. Whether the current node is added to the output before, after, or between its children doesn't make a difference - the algorithms walk the nodes of the tree in exactly the same order.
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.
How can you convert Binary Tree to Binary Search Tree with O(1) extra space ?
Converting an unordered binary tree into an ordered binary search tree is trivial, but a bit more difficult to do fast.
Here's a naive implementation that should satisfy your criteria, I will not describe the actual steps to take, just the overall algorithm.
Grab a random leaf node from your existing tree
Unlink the leaf node from your existing tree
Make the node the root of your new binary search tree
Grab another random leaf node from your existing tree
Unlink that node from your existing tree
Find the right spot for, and link the node, into your new binary search tree
Repeat step 4-6 until the original tree is empty
You should require only a few variables, like the parent of the leaf node you're unlinking (unless the nodes has parent-links), the root node of the new tree, and a couple of temporary variables, all within your O(1) space criteria.
This will not produce an optimal binary search tree. For that you need to either sort the nodes before adding them, and adding them in the right order, or use a balancing binary search tree, like a red-black tree or a splay tree.
Convert Binary Tree to a doubly linked list- can be done inplace in O(n)
Then sort it using merge sort, nlogn
Convert the list back to a tree - O(n)
Simple nlogn solution.