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The number of branches is 8.

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Q: What is the Number of possible binary trees with 3 nodes?
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Number of all possible binary trees with 2 nodes is?

Two: 1. root and left child 2. root and right child


There are 8 15 13 14 nodes were there in 4 different trees Which of them could have formed a full binary tree?

In general: There are 2n-1 nodes in a full binary tree. By the method of elimination: Full binary trees contain odd number of nodes. So there cannot be full binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete binary tree but not a full binary tree. So the correct answer is 15. niraj


Why you represent the general trees as binary trees?

General trees are not binary trees. It is the other way around, however, see the last paragraph for a different answer - explanation first... A binary tree is one with two possible child nodes, a left node and a right node, either of which might be not present. This particular representation implies a certain order between the node and its children, and if you walk the tree from bottom left to bottom right, you will traverse the nodes in order. A general tree is one with any number of possible child nodes, including no child nodes, so a binary tree is an example of a general tree, while a general tree is a generalization of a binary tree. However, in the general tree, the meaning of the child nodes might not have any specific ordering, like those in a binary tree, unless the general tree has other information contained in the node about order, because the concept of left and right has no implied meaning when there are more than two children. But, as promised, if the general tree has order, it is always possible to represent the general tree as a binary tree - there will just be more nodes, but they will only contain zero, one, or two children, and they will have an implied order.


Methods for storing binary trees?

Nodes, references and arrays are the methods for storing binary trees. It can also be stored in breath first order.


How many different trees are possible with 10 nodes?

1014 it is. no of different trees possible with n nodes is (2^n)-n thanx


How many ordered trees are possible with 3 nodes?

6


How many no of ordered trees are possible with 3 nodes?

12


How many different binary trees can be made from three nodes?

As far as i Know, just one.Do you know any formula to calculate how many binary search trees are possible?--answer:(2n C n) / (n+1) = ( factorial (2n) / factorial (n) * factorial (2n - n) ) / ( n + 1 )where 'n' is number of element (integer/string)like:N Number of BST1 12 23 54 145 426 132and so on


What is the difference between extended binary tree and a binary search tree?

A strictly binary tree is one where every node other than the leaves has exactly 2 child nodes. Such trees are also known as 2-trees or full binary trees. An extended binary tree is a tree that has been transformed into a full binary tree. This transformation is achieved by inserting special "external" nodes such that every "internal" node has exactly two children.


How many binary trees are possible with 3 nodes?

42http://en.wikipedia.org/wiki/Catalan_number


What is NULL branches in trees?

In a decision tree data structure, a NULL branch is a decision that has no outcome. The maximum possible number of null branches is equal to n + 1, where n is the number of nodes.


How do you count all structurally different possible Binary Trees?

please tell me answer of this question. Suppose you are building an N node binary search tree with the values 1...N. how many structurally different binary trees is there that store those values? write a recursive function that, gives the number of distinct values, computes the number of structurally unique binary search trees that store those values. For example, countTrees(4) should return 14, since there are 14 structurally unique binary search trees that store 1,2,3 and 4. The base case us easy, and the recursion is short but dense. your code should not construct any actual trees; it's just a counting problem.