if a full binary tree has 36 internal vertices, how many leaves does it have?
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if a full binary tree has 36 internal vertices, how many leaves does it have?
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- D) Draw a binary tree whose inorder traverse is T , W , B , P , Y , R , M , X , L , K , S , A and preorder traverse is R , T , P , W , B , Y , K , M , L , X , S , A1. How many different binary trees are possible with n nodes? Solution: For example, consider a tree with 3 nodes (n=3). It will have a maximum combination of 5 different trees. 2. Assume that each node has either 0 or 2 children. Given string: I L I L L a. Give the pre order traversal. b. Give the Postorder traversalIs the following statement true? Justify your answer. Let B be a binary tree. If for each of vertices (v) a data item is inserted into vertex (v) and they are bigger than the data item inserted in the left son of vertex (v) and smaller at the same time, than data item inserted in the right son of the vertex (v). B is a search tree. Definition of data item(data entry): Data item is one unit of data in a storage record.
- Algorithm of spanning tree 1. Implement the Kruskal's algorithm of spanning tree 2. Write comments in each line as you understand3. Screenshots of three times output, for 5 verteces, 7 verteces, 8 verteces=============================================================/// Implementation of Kruskal"s algorithm on Spanning Tree#include<stdio.h>#define max 99int k;int array[8];int u,v;void main(){ int vertex, edge =1; /// ***We initialized but didn't increment no where int i,j, p,q; int edgecost[8][8]; int min = max, totalmincost = 0; /// To find out the minimum and total of those printf("\nEnter the quantity of Verteces: "); scanf("%d", &vertex); /// 1,2 -> 0 printf("\nEnter the cost values in matrix:\n"); for(i=1; i<=vertex; i++) { printf("\n"); for(j=1; j<=vertex; j++) { printf("cost[%d][%d]: ", i,j); scanf("%d", &edgecost[i][j]); if(edgecost[i][j] == 0)…Create the Sample class, which has a constructor that accepts an array p[] of double values as an input and supports the following two operations: Return an index i with a probability of p[i]/T (where T is the total of the numbers in p[]) and alter(i, v) to change the value of p[i] to v. Use a full binary tree with an inferred weight of p[i] for each node. Keep the total weight of all the nodes in its subtree in each node. Pick a random number between 0 and T to obtain a random index and use the cumulative weights to select which branch of the subtree to examine. Change the weights of all nodes on the path from the root to i when updating p[i]. Avoid explicit pointers, as we do for heaps.Draw a binary search tree {15, 5, 20, 70, 3, 10, 60, 90, 16} Reverse path of Binary Search Tree drawn
- Suppose that instead of each node x keeping the attribute x.p, pointing to x’s parent, it keeps x.succ, pointing to x’s successor. Give pseudocode for SEARCH, INSERT, DELETE, and EDIT routine on a binary search tree T using this representation. These procedures should operate in time O(h), where h is the height of the tree T. (Hint: You may wish to implement a subroutine that returns the parent of a node.) Show these types of pseudocode using java. Data structures and algorithmsIn DrRacket: Define the difference between Binary Search Trees and Nodes? A) A Binary Search Tree is a Node, but a Node isn’t always a Binary Search Tree. B) A Node is a Binary Search Tree, but a Binary Search Tree isn’t always a Node. C) Binary Search Trees may contain Nodes, but Nodes cannot contain Binary Search Trees. D) Nodes may contain Binary Search Trees, but Binary Search Trees cannot contain Nodes. E) They are the same thingAnswer the following questions about treeB. a. What is the height of the tree? b. Which nodes are on level 3? c. Which levels have the maximum number of nodes that they could contain? d. What is the maximum height of a binary search tree containing these nodes? Draw such a tree. e. What is the minimum height of a binary search tree containing these nodes? Draw such a tree. f. What is the order in which the nodes are visited by an ignorer traversal? g. What is the order in which the nodes are visited by a preorder traversal? h. What is the order in which the nodes are visited by a postorder traversal?
- A mathematician applies Prim’s Algorithm to find a minimum spanning tree for the weighted graph starting at vertex D. The order of the edges picked so far is AD, AB, BE, and EF. The next edge selected when applying Prim’s Algorithm should be BF FG CG CE What is the total weight of the tree that uses the edges DE, CE, CG, FG and BF? 93 92 91 90 The tree described in problem #8 is a spanning True False8_ I need completion of code. Given a binary tree, check whether it is a mirror ofitself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \3 4 4 3But the following [1,2,2,null,3,null,3] is not: 1 / \ 2 2 \ \ 3 3Note:Bonus points if you could solve it both recursively and iteratively.""" # TC: O(b) SC: O(log n)def is_symmetric(root): if root is None: return True return helper(root.left, root.right) def helper(p, q): if p is None and q is None: return True if p is not None or q is not None or q.val != p.val: return False return helper(p.left, q.right) and helper(p.right, q.left) def is_symmetric_iterative(root): if root is None: return True stack = [[root.left, root.right]] while stack: left, right = stack.pop() # popleft if left is None and right is None: continue. .Discussion:1. Depth-first search (DFS) is a technique that is used to traverse a tree or a graph.DSF technique starts with a root node and then traverses the adjacent nodes ofthe root node by going deeper into the graph. In the DFS technique, the nodes aretraversed depth-wise until there are no more children to explore.- Once we reach the leaf node (no more child nodes), the DFS backtracks andstarts with other nodes and carries out traversal n a similar manner. DFStechnique uses a stack data structure to store the nodes that are beingtraversed. DFS Technique (Depth-First Traversal)o Following is the algorithm for the DFS technique.o Algorithm:1. Start with the root node and insert it into the stack2. Pop the item from the stack and insert into the ‘visited’ list3. For the node marked as ‘visited’ (or in visited list), add the adjacent nodesof this node that are not yet marked visited to the stack.4. Repeat steps 2 and 3 until the stack is empty.