I know well about Full Binary Tree and Complete Binary Tree. But unable to make Full binary tree with only 6 nodes.
The answer is No. You can't make a Full binary tree with just 6 nodes. As the definition in the Wikipedia says:
A full binary tree (sometimes referred to as a proper or plane
binary tree) is a tree in which every node has either 0 or 2
children. Another way of defining a full binary tree is a recursive
definition. A full binary tree is either:
A single vertex.
A tree whose root node has two subtrees, both of which are full binary trees.
Another interesting property I noticed is that, the number of nodes required to make a full binary tree will always be odd.
Another way to see that a full binary tree has an odd number of nodes:
Starting with the definition of a full binary tree (Wikipedia):
a tree in which every node has either 0 or 2 children.
This means that the total number of child nodes is even (0+2+2+0+...+2 is always even). There is only one node that is not a child of another, which is the root. So considering that node as well, the total becomes odd.
By consequence there is no full binary tree with 6 nodes.
Elaborating on #vivek_23's answer, this is, unfortunately, not possible. There's a beautiful theorem that says the following:
Theorem: Any full binary tree has 2L - 1 nodes, where L is the number of leaf nodes in the tree.
The intuition behind this theorem is actually pretty simple. Imagine you take a complete binary tree and delete all the internal nodes from it. You now have a forest of L single-node full binary trees, one for each leaf. Now, add the internal nodes back one at a time. Each time you do, you'll be taking two different trees in the forest and combining them into a single tree, which decreases the number of trees in the forest by one. This means that you have to have exactly L - 1 internal nodes, since if you had any fewer you wouldn't be able to join together all the trees in the forest, and if you had any more you'd run out of trees to combine.
The fact that there are 2L - 1 total nodes in a full binary tree means that the number of nodes in a full binary tree is always odd, so you can't create a full binary tree with 6 nodes. However, you can create a full binary tree with any number of odd nodes - can you figure out how to prove that?
Hope this helps!
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
(Binary search tree is a binary tree where each node can have 2 children atmost, the right being larger than the node, and the left should be smaller than the node.)
I have a theory that i want to disprove. It says that for any binary tree, the if we take a search path (call it S) to a leaf node, then any node on the LEFT of S must be smaller than any node on S, and any node of the RIGHT must be larger than any node on S. In other words: node on left < node on S < node on right. Is there any counter-example to disprove this theory?
For example if we have this tree:
A search path for node K would be M->F->H->K
The set of nodes on the left contains C, A, D, G
The set on the right contains V,S,P,T,X,W
What is a good counter example?
Thank you.
This isn't really an answer, but it wouldn't fit in a comment...
I think your definition of "binary search tree" is a bit lacking - after all, this would meet your definition:
B
\
C
/ \
A D
However, that's not a true binary search tree - your definition lacks the recursive relationship. In a binary search tree, all elements in the left subtree of a node are less than the node label, and all elements in the right subtree are greater - not just the immediate children.
Perhaps having a more precise definition would help you in thinking about your "theory".
I know that from a general tree you can construct a unique binary tree, but is the reverse true? i.e can you get a unique general tree from a binary tree?
Yes. The following transformation is reversible:
Given a general tree with ordered but not indexed children,
encode the first child as the left child of its parent, and each other node as a right child of its (former) sibling.
The reverse is:
Given a binary tree with distinguished left and right children, read the left child of a node as its first child and the right child as its next sibling.
So, the following tree
a
/|\
b c d
is encoded as
a
/
b
\
c
\
d
while the following tree
a
/ \
b c
|
d
is encoded as
a
/
b
/ \
d c
(read: d is the first child of b, c is the sibling of a).
Note that you can encode any rooted forest (with ordered components, otherwise the representation is not unique) by assigning a sibling to the root, so this
a
/ \
b c
\ \
d e
would be read as
a c e
/ \
b d
here is another method to get a unique general (undirected) tree from a binary tree:
a vertex binary tree may have 0...3 graph neighbors.
append 12 nodes to the root
append 8 nodes to each left child
append 4 nodes to each right child
this operation is reversible:
label the node with at least 12 neighbors "root". If not unique, fail.
label each node with 8..11 neighbors "left".
label each node with 4..7 neighbors "right".
remove all leaves
orient all edges away from the root
if any node has more than one left child or more than one right child, fail.
So,
There is a bijection between ordered rooted trees and binary trees (first and second algorithm).
Since any general tree can be arbitrarily rooted, there is a injection from general (directed or undirected) trees to binary trees.
There is an injection from binary trees to general undirected trees (third algorithm)
Since there is an injection from binary trees to general trees and back, there must exist a bijection between general (directed or undirected) trees and binary trees.
I feel unlikely. Usually, binary tree distinguishes left child and right child. However, general trees don't.
How are we supposed to get a unique general tree from these two binary tree.
X X
/ \ / \
Y Z Z Y
And how about these two?
X X
/ \
Y Y
On the other hand,
If you choose to not distinguish left or right child of a binary tree, or choose to respect the sequence children appears in a general tree, just map each binary tree to itself. That will be a unique general tree for each binary tree.
Can anyone please explain the difference between binary tree and binary search tree with an example?
Binary tree: Tree where each node has up to two leaves
1
/ \
2 3
Binary search tree: Used for searching. A binary tree where the left child contains only nodes with values less than the parent node, and where the right child only contains nodes with values greater than or equal to the parent.
2
/ \
1 3
Binary Tree is a specialized form of tree with two child (left child and right Child).
It is simply representation of data in Tree structure
Binary Search Tree (BST) is a special type of Binary Tree that follows following condition:
left child node is smaller than its parent Node
right child node is greater than its parent Node
A binary tree is made of nodes, where each node contains a "left" pointer, a "right" pointer, and a data element. The "root" pointer points to the topmost node in the tree. The left and right pointers recursively point to smaller "subtrees" on either side. A null pointer represents a binary tree with no elements -- the empty tree. The formal recursive definition is: a binary tree is either empty (represented by a null pointer), or is made of a single node, where the left and right pointers (recursive definition ahead) each point to a binary tree.
A binary search tree (BST) or "ordered binary tree" is a type of binary tree where the nodes are arranged in order: for each node, all elements in its left subtree are less to the node (<), and all the elements in its right subtree are greater than the node (>).
5
/ \
3 6
/ \ \
1 4 9
The tree shown above is a binary search tree -- the "root" node is a 5, and its left subtree nodes (1, 3, 4) are < 5, and its right subtree nodes (6, 9) are > 5. Recursively, each of the subtrees must also obey the binary search tree constraint: in the (1, 3, 4) subtree, the 3 is the root, the 1 < 3 and 4 > 3.
Watch out for the exact wording in the problems -- a "binary search tree" is different from a "binary tree".
As everybody above has explained about the difference between binary tree and binary search tree, i am just adding how to test whether the given binary tree is binary search tree.
boolean b = new Sample().isBinarySearchTree(n1, Integer.MIN_VALUE, Integer.MAX_VALUE);
.......
.......
.......
public boolean isBinarySearchTree(TreeNode node, int min, int max)
{
if(node == null)
{
return true;
}
boolean left = isBinarySearchTree(node.getLeft(), min, node.getValue());
boolean right = isBinarySearchTree(node.getRight(), node.getValue(), max);
return left && right && (node.getValue()<max) && (node.getValue()>=min);
}
Hope it will help you. Sorry if i am diverting from the topic as i felt it's worth mentioning this here.
Binary Tree stands for a data structure which is made up of nodes that can only have two children references.
Binary Search Tree (BST) on the other hand, is a special form of Binary Tree data structure where each node has a comparable value, and smaller valued children attached to left and larger valued children attached to the right.
Thus, all BST's are Binary Tree however only some Binary Tree's may be also BST. Notify that BST is a subset of Binary Tree.
So, Binary Tree is more of a general data-structure than Binary Search Tree. And also you have to notify that Binary Search Tree is a sorted tree whereas there is no such set of rules for generic Binary Tree.
Binary Tree
A Binary Tree which is not a BST;
5
/ \
/ \
9 2
/ \ / \
15 17 19 21
Binary Search Tree (sorted Tree)
A Binary Search Tree which is also a Binary Tree;
50
/ \
/ \
25 75
/ \ / \
20 30 70 80
Binary Search Tree Node property
Also notify that for any parent node in the BST;
All the left nodes have smaller value than the value of the parent node. In the upper example, the nodes with values { 20, 25, 30 } which are all located on the left (left descendants) of 50, are smaller than 50.
All the right nodes have greater value than the value of the parent node. In the upper example, the nodes with values { 70, 75, 80 } which are all located on the right (right descendants) of 50, are greater than 50.
There is no such a rule for Binary Tree Node. The only rule for Binary Tree Node is having two childrens so it self-explains itself that why called binary.
A binary search tree is a special kind of binary tree which exhibits the following property: for any node n, every descendant node's value in the left subtree of n is less than the value of n, and every descendant node's value in the right subtree is greater than the value of n.
Binary tree
Binary tree can be anything which has 2 child and 1 parent. It can be implemented as linked list or array, or with your custom API. Once you start to add more specific rules into it, it becomes more specialized tree. Most common known implementation is that, add smaller nodes on left and larger ones on right.
For example, a labeled binary tree of size 9 and height 3, with a root node whose value is 2. Tree is unbalanced and not sorted.
https://en.wikipedia.org/wiki/Binary_tree
For example, in the tree on the left, A has the 6 children {B,C,D,E,F,G}. It can be converted into the binary tree on the right.
Binary Search
Binary Search is technique/algorithm which is used to find specific item on node chain. Binary search works on sorted arrays.
Binary search compares the target value to the middle element of the array; if they are unequal, the half in which the target cannot lie is eliminated and the search continues on the remaining half until it is successful or the remaining half is empty. https://en.wikipedia.org/wiki/Binary_search_algorithm
A tree representing binary search. The array being searched here is [20, 30, 40, 50, 90, 100], and the target value is 40.
Binary search tree
This is one of the implementations of binary tree. This is specialized for searching.
Binary search tree and B-tree data structures are based on binary search.
Binary search trees (BST), sometimes called ordered or sorted binary trees, are a particular type of container: data structures that store "items" (such as numbers, names etc.) in memory. https://en.wikipedia.org/wiki/Binary_search_tree
A binary search tree of size 9 and depth 3, with 8 at the root. The leaves are not drawn.
And finally great schema for performance comparison of well-known data-structures and algorithms applied:
Image taken from Algorithms (4th Edition)
Binary search tree: when inorder traversal is made on binary tree, you get sorted values of inserted items
Binary tree: no sorted order is found in any kind of traversal
A binary tree is a tree whose children are never more than two. A binary search tree follows the invariant that the left child should have a smaller value than the root node's key, while the right child should have a greater value than the root node's key.
To check wheather or not a given Binary Tree is Binary Search Tree here's is an Alternative Approach .
Traverse Tree In Inorder Fashion (i.e. Left Child --> Parent --> Right Child ) ,
Store Traversed Node Data in a temporary Variable lets say temp , just before storing into temp , Check wheather current Node's data is higher then previous one or not .
Then just break it out , Tree is not Binary Search Tree else traverse untill end.
Below is an example with Java:
public static boolean isBinarySearchTree(Tree root)
{
if(root==null)
return false;
isBinarySearchTree(root.left);
if(tree.data<temp)
return false;
else
temp=tree.data;
isBinarySearchTree(root.right);
return true;
}
Maintain temp variable outside
A tree can be called as a binary tree if and only if the maximum number of children of any of the nodes is two.
A tree can be called as a binary search tree if and only if the maximum number of children of any of the nodes is two and the left child is always smaller than the right child.
In a Binary search tree, all the nodes are arranged in a specific order - nodes to the left of a root node have a smaller value than its root, and all the nodes to the right of a node have values greater than the value of the root.
In a binary tree, each node has 2 child nodes the left node and the right node.
A Binary Search tree is a special kind of tree in which the nodes are sorted, the left node is smaller than the parent node and the left node is bigger than the parent node.
The binary tree allows duplicate values, Binary search tree doesn't allow duplicate values also carrying out any kind of operation is faster in Binary search tree than in Binary tree since BST is sorted
A binary tree is a tree, in which each node can have at most 2 children.
A binary search tree is a further modification of this, giving a certain relationship to the parent and the two children. Since, there are only two children, i.e., left and right child; the relation is defined as follows:
Left Child <= Parent <= Right Child
It is actually, that simple.