→ Pre order traverses Root, Left, Right.
→ Post order traverses Left, Right, Root.
→ In-order traverses Left, Root, Right.
Tree-1: Post order: GJIHEFBDCA
Tree-2: In order: GJIHEFBDCA

Step-1: Post order traversals yields from left,right and root.

Step-2: The post order sequence is 4526731. The same we have to take it for above constraint.
Step-3: Then the sequence will be ab–cd*+ because 1= +, 2= -, 3= *, 4= a, 5=b, 6=c and 7=d.

In a Max. Binary Heap, the key value at each node is as least as large as the values at its children. Similarly in Min Binary Heap, the key at root must be minimum among all keys present in Binary Heap.

A Binary Tree is full if every node has 0 or 2 children. So, in such case, the binary tree with n leaves contains a total of 2*n-1 nodes.

→ A binary tree where each non-leaf node has exactly two children. The number of leaves is n+1. Total number of nodes is 2n+1.
Note: if two leaves having the same parent are removed from the tree, the number of leaves and non-leaves will decrease by 1 because the parent will be a leaf. That means the difference between them is constant. And for a tree having just one node that difference is 1.

A full binary tree is a tree in which every node other than the leaves has two children. In short, a full binary tree with N leaves contains 2N - 1 nodes.

For a given nodes”n” , we can get n^n-2 number of different trees
n=4. Here, n is number of nodes.
4^4-2=16.
Given options are wrong options.
  Note: Correct answer is 16

● A full binary tree (sometimes proper binary tree or 2-tree) is a tree in which every node other than the leaves has two children
● In a binary tree - a tree where each non-leaf node has exactly two sons - number of leaves is n+1. Total number of nodes is 2n+1.
Full Binary Tree:

In the above tree,non-leaf nodes are 7 so total nodes are 2x7+1=15

It is a self balancing tree with height difference at most 1. A balance factor in AVL tree is used to check what rotation to make.

In order:​ The left subtree is visited first, then the root and later the right subtree.
Pre order:​ The root node is visited first, then the left subtree and finally the right subtree.
Post order:​ First we traverse the left subtree, then the right subtree and finally the root node.

B trees need not be the binary tree. B trees may have more than 2 children. The order of B tree is maximum number of children a node can have.

The number of nodes is equal to 2​ ^k​ -1 where k≥1. So the minimum level is 1.
For example

The minimum number of nodes 2​ ²​ -1=3

Queue data structures is used in level order traversal of a binary tree.
Example:
Level order traversal of binary tree.
For each node, first the node is visited and then it’s child nodes are put in a FIFO queue.
printLevelorder(tree)
1) Create an empty queue q
2) temp_node = root /*start from root*/
3) Loop while temp_node is not NULL
a) print temp_node->data
b) Enqueue temp_node’s children (first left then right children) to q
c) Dequeue a node from q and assign it’s value to temp_node
Note:
Level order traversal of binary tree is time Complexity is O(n) where n is number of nodes in the binary tree.

→ Height of the full binary tree 2^h -1.
Full binary tree consists of either 0 or 2 childs . So one child to the node is not possible.
The number of nodes with the degree of 1 is 0.
The nodes with degree 0 are leaf nodes, 2^h-1 = 2⁹ = 512.
The nodes with degree 2 are internal nodes 2^h-1 - 1 = 511.

● The number of nodes n in a full binary tree, is at least n = 2 h + 1 and at most n = 2 ^h+1 − 1 , where h is the height of the tree. A tree consisting of only a root node has a height of 0.
● The number of leaf nodes l in a perfect binary tree, is l = ( n + 1 )/2 because the number of non-leaf (a.k.a. internal) nodes

● This means that a perfect binary tree with l leaves has n = 2 l − 1 nodes. ● In a balanced full binary tree, h = ⎡ log ₂ (l)⎤ + 1 = ⎡ log ₂ ((n + 1 )/2)⎤ + 1 = ⎡ log ₂ (n + 1 )⎤
● In a perfect full binary tree, l = 2 ^h thus n = 2 ^h+1 − 1
● The maximum possible number of null links (i.e., absent children of the nodes) in a complete binary tree of n nodes is ( n + 1 ) , where only 1 node exists in bottom-most level to the far left.
●The number of internal nodes in a complete binary tree of n nodes is ⎣ n/2⎦ .

There is one binary tree with one node. There are two differently shaped trees with two nodes. There are 14 different (shaped) binary trees with four nodes. These different trees are shown below.

Preorder traversal yields (Root, left, right)
Option D:
Let draw binary search tree for the given sequence,

After traversing through this tree we will get same sequence.

A binary tree is a tree data structure in which each node has at most two children, which are referred to as the left child and the right child

The number of decisions or, alternatively, the number of branches traversed, in reaching a node is called the path length or path to the node.

Here, inoperate the traversal yields the expression means INORDER.
Inorder→ left,root and right.
a+b*d-c/f

→A full binary tree (sometimes proper binary tree or 2-tree) is a tree in which every node other than the leaves has two children.
→The height of the binary tree is the longest path from root node to any leaf node in the tree

→ They are given definition of full binary tree. The full binary tree each node has exactly either zero or two children.
→ The maximum height of the tree will be (n–1) / 2. It is standard property of full binary tree.

Inorder: Left root Right
Pre order: Root Left Right
Post order: Left Right Root
Inorder: d b e a f c g

Pre order: a b d e c f g
Post order: d e b f g c a

A splay tree is a self-balancing binary search tree with the additional property that recently accessed elements are quick to access again. It performs basic operations such as insertion, look-up and removal in O(log n) amortized time. For many sequences of non-random operations, splay trees perform better than other search trees, even when the specific pattern of the sequence is unknown.

Inorder traversal technique in binary search tree will always produce ascending order.
Example:

In this traversal method, the left subtree is visited first, then the root and later the right subtree. We should always remember that every node may represent a subtree itself.

a, c, d are going to unbalance.

Binary search tree is an example of divide and conquer technique.
Procedure:
1. Divide: select lower or upper half
2. Conquer: search selected half
3. Combine: None
Recurrence relation​ : T(n)=T(n/2)+Θ(1)
Using masters theorem,
a=1, b=2, k=0,p=0
T(n)=Θ(logn)

a, c, d are going to unbalance.

A Binary Tree is full if every node has 0 or 2 children. So, in such case, the binary tree with n leaves contains a total of 2*n-1 nodes.

There are 4 leaf nodes in the above tree.
So, total number of nodes are 2*4-1
=8-1
=7

Binary tree is a tree where each node has at most 2 children nodes.
Properties:
→The number of nodes at depth d in a perfect binary tree = 2^d
→A perfect binary tree of height h has 2^h+1 -1 nodes
→Number of leaf nodes in a perfect binary tree of height h = 2^h
→Number of internal nodes in a perfect binary tree of height h = 2^h -1
→The minimum number of nodes in a binary tree of height h = h+1
→The maximum number of nodes in a binary tree of height h = 2^h+1 -1

Step-1: Post order traversal is Left, Right and Root according to post order we are dividing into in-order list.

Step-2:

Step-3: Final tree is

Step-4: Preorder traversal is Root,left and right
preorder= a b e j k n p o f d g l c m i h
Note: Given options are wrong. Excluded for evaluation.

AVL tree is a self balancing tree with height difference at most 1. A balance factor in AVL tree is used to check what rotation to make.

→ AVL tree is a self balancing tree with height difference at most 1. A balance factor in AVL tree is used to check what rotation to make.
→ ALV tree

Eccentricity of a vertex u, is
E(u) = max_u∈v(G) d(u, v)
i.e., the maximum distance of a node to other nodes is the eccentricity of that node

Radius: The minimum eccentricity from all the vertices is considered as the radius of the Graph G denoted by r(G).
Here radius of the graph, r(G)=3
Center:- If the eccentricity of a graph is equal to its radius, then it is known as the central point of the graph.
And in above graph vertex "h" and " c" have eccentricity equal to the radius of the graph. So vertex h and c are the center of graph.
Hence correct answer is option (D).

Post order traversal tree is look like

Postorder: DEBFCA
Inorder: DBEAFC
Preorder: ABDECF

Binary search tree properties: 1. All items in the left subtree are less than root
2. All items in the right subtree are greater than or equal to the root
3. Each subtree is itself a binary search tree

Double-Ended Queue is a data structure in which insertion and deletion of data takes place from different ends. The insertion of data takes place from REAR end and deletion of data takes place from FRONT end. It follows FIFO( First In First Out) order for implementation.

Priority Queue is like a regular queue or stack data structure, but here each element has a "priority" associated with it. In a priority queue, an element with high priority is served before an element with low priority.
Tree: A tree is a data structure containing a collection of nodes. Here a node can have any number of children. A node with no child is called as leaf node and a node with 1 or more child is called parent node. Each parent node have a pointer to its children.

Post order traversal tree is look like

Postorder: DEBFCA
Inorder: DBEAFC
Preorder: ABDECF

The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to 2⁰ + 2¹ + ….. 2^h. because every phase we are getting 2^h.
Suppose, in level-3, it have 2³=8 nodes.

Given data
-- nodes=8
-- Total number of trees=?
Possibility-1: To find total number of trees using n^n-2.
= 8^8-2
= 262144 trees are possible.
Possibility-2: According to given problem, they are using binary tree based on below formula
= 2ⁿ -n

= 2⁸ -8
= 248
Note: They are not given proper input.

According to binary two most distinct nodes are in
level-1: 1 and 2
level-2: 3 and 6.
[Note: Assume that root starts from level-0]

The height of the binary tree is O(log₂n). As per the given constraint, we have to calculate individually of left farthest node and right farthest node height.
=log₂n*log₂n
=2*log₂n or (log₂n)²

→ In a full binary tree of height k, there are 2k-1 internal nodes.
→ A Binary Tree is full if every node has 0 or 2 children.
→ In a Full Binary, number of leaf nodes is number of internal nodes plus 1
L = I + 1
Where,
L = Number of leaf nodes,
I = Number of internal nodes.

A heap is a complete binary tree. A binary tree is said to have heap property if the elements along any path from root to leaf are non–increasing

Binary search tree will follow some rules
1. Left subtree is smaller than his root node
2. Right subtree is greater than or equal to his root
Based on above rules, the search operation will perform

The height of a binary tree with ‘n’ nodes, in the worst case is O(n). Sometimes, it will get left skewed/ right skewed based on given input.

→ The depth of binary tree starts from 0 but not 1.
→ To find maximum number of nodes in a binary tree of depth using formula is
= 2^h+1 -1
= 2¹¹ -1
= 2048-1
= 2047

→ A node's path length is the number of links (or branches) required to get back to the root.
→ The root has path length zero and the maximum path length in a tree is called the tree's height.
→ The sum of the path lengths of a tree's internal nodes is called the internal path length and the sum of the path lengths of a tree's external nodes is called the external path length.
External Path Length:

The sum over all external (square) nodes of the lengths of the paths from the root of an extended binary tree to each node. For example, in the tree above, the external path length is 25 (Knuth 1997, pp. 399-400). The internal and external path lengths are related by
E = I + 2n,
where n is the number of internal nodes.

Using catalan formula to find the total number of binary trees is (2n)! / (n! * (n+1)!)

Here, n=6

= (2*6)! / (6! * (6+1)!)

= (12)! / (6! * (7)!)

= 479001600 / (720*5040)

= 132