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Java - The node (80, 85, 90) is split. Enter each node's keys after the split, or "none" if the node doesn't exist.
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- 1) In a tree, an interior node is a node that has no children. a. True b. False 2) In a tree, the root item has no parent item. a. True b. Falseclass BST: def __init__(self, root = None): if root != None: root.setparent(None) self.__root = root def getroot(self): # Returns the root node of the tree ####################################################################### # Remove the pass and write your code ####################################################################### pass ####################################################################### # End code ####################################################################### def setroot(self, node): # Sets the root node of the tree ####################################################################### # Remove the pass and write your code ####################################################################### pass ####################################################################### # End code…Part 1: create random binary trees For each tree, create a tree according to the following procedure. Toss a fair coin up to 100 tossesStart from the root node as current node. A: Perform a toss: determine whether the current node will have at least one descendent. If Head, there is at least one descendent. Then go to B. If Tail, there is no descendent. Then go to D. B: Perform a toss: determine whether the current node will have one or two children. If Head, the current node will have one child. Then go to C.If Tail, the current node will have two children. Assign two random integer values to the two children. Then go to D. C: Perform a toss: determine whether the current node will have left or right child.If Head, the current node will have a left child. Assign a random integer value to the child. Then go to D.If Tail, the current node will have a right child. Assign a random integer value to the child. Then go to D. D: move the current node to the next node (using…
- int doo(node<int>*root){ if(root !=0 ) { if(root->left==0 && root->right==0) return root->data; int L=doo(root->left); int R=doo(root->right); if(L>R) return R; else return L;} } this code used for a. find the sum of leaves items in a binary tree b. find the minimum item in a binary tree c. find the minimum item in the leaves of a binary tree d. the maximum item in the leaves of a binary treeHuffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only the leaves will contain a letter and its frequency count. All other nodes will contain a null instead of a character, and the count of the frequency of all of it and its descendant characters. For instance, consider the string ABRACADABRA. There are a total of characters in the string. This number should match the count in the ultimately determined root of the tree. Our frequencies are and . The two smallest frequencies are for and , both equal to , so we'll create a tree with them. The root node will contain the sum of the counts of its descendants, in this case . The left node will be the…int doo(node<int>*root){ if(root !=0 ) { if(root->left==0 && root->right==0) return root->data; return doo(root->right); } } this code used for a. find the maximum item in a binary search tree if it has a right sub tree b. find the number of right leaves in a binary tree if it has a right sub tree c. find the maximum item in a binary tree if it has a right sub tree d. find the first right item in the binary tree if it has a right sub tree
- Please draw the following:Form a binary tree with the root node holding 20, the left and right child of the node holding 20 respectively holding 75 and 43. The next level nodes hold the data 84, 90, 57 and 71 from left to right. The node holding 84 has left and right children that respectively hold 96 and 91 and the node holding 90 has a left child that holds 93.Implement a Binary Search Tree (BST). Implement the following tree operations: a. new (create a new tree) b. isEmpty (determine if the BST is empty. Will display either True or False) c. isLeaf (determine if the node is a leaf. Will display either True or False) d. path (will display the path in BST from a root to the node. If the node is present it will will display the complete path otherwise will display "Null") e. search (will search the BST for a key/value. if the key is present it will will display "Present" otherwise will display "Absent") f. insert (will insert a key as a node in the BST; acts as root, left and right from the prior activity) g. delete (will delete a key from the BST) h. transversal (will display all the nodes in a BST; Use either Breadth or Inorder transversal from the prior activity)Postorder: 7 6 8 5 2 9 4 10 3 1 Inorder: 6 7 2 5 8 1 4 9 3 10 After drawing the tree, at what level does node 10 belongs? (Numeric Answer ONLY)
- In Java, True or False: Every binary tree has at least one node. TRUE FALSE Every non-empty tree has exactly one root node. TRUE FALSE Every node in a binary tree has at most one child. TRUE FALSE Every non-root node in a binary tree has exactly two parents. TRUE FALSERefer to the tree on the picture and answer the following questions 1. Aside from its children, list 2 other descendants of node C. Your answer must be values that are adjacent when the nodes are arranged alphabetically in ascending order. For example, the descendant nodes are: ABX Y. You cannot answer A X since you have skipped B and B Y since X has been skipped. * 2. What is the path from node A to node L? * 3. What is the length of the path from node C to node N? Note: answer with the number's symbol and not the word form of the number.* 4. List 2 nodes that are at level 2 of the tree. Your answer must be values that 1 point are adjacent when the nodes are arranged alphabetically in ascending order. For example, the nodes at a certain level are: A B X Y. You cannot answer A X since you have skipped B and B Y since X has been skipped. * 5. What is the height of the tree? Note: answer with the number's symbol and not the word form of the number.* 6. What is the height of node C? Note:…Postorder: 7 6 8 5 2 9 4 10 3 1 Inorder: 6 7 2 5 8 1 4 9 3 10 After drawing the tree, what/who is the parent node of node 8? (Numeric Answer ONLY)
