Explanation: summing up every left leaf in the tree gives us: 9 + 15 = 24
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- Java / Trees: *Please refer to attached image* How many leaf nodes are there on this tree? Multiple chocie. 0 3 4 10e) (2a + 5b) ^ 3 * (x - 7y) ^ 4 f) Evaluate the binary expression tree in e) usingthe arrows method. Assume a=1, b=0, x=2, y=0BOX IN YOUR ANSWERclass 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…
- Please help! // TreeNode.java class TreeNode<T> { T item; TreeNode<T> leftChild; TreeNode<T> rightChild; public TreeNode(T newItem) { // Initializes tree node with item and no children. item = newItem; leftChild = null; rightChild = null; } // end constructor public TreeNode(T newItem, TreeNode<T> left, TreeNode<T> right) { // Initializes tree node with item and // the left and right children references. item = newItem; leftChild = left; rightChild = right; } // end constructor } // end TreeNode // BinaryTreeBasis.java public abstract class BinaryTreeBasis<T> { protected TreeNode<T> root; public BinaryTreeBasis() { root = null; } // end default constructor public BinaryTreeBasis(T rootItem) { root = new TreeNode<T>(rootItem, null, null); } // end constructor public boolean isEmpty() { // Returns true if the tree is empty, else returns false. return root == null; } // end isEmpty public void makeEmpty() { // Removes all nodes from the…Write the method "isALeafNode" to check weather a specific node is a leaf or not { }(public boolean isALeafNode(Node nWrite the following string with your information: string=”your fullName 19-arid-yourNo” //e.g.Isha Tir Razia 19-ARID-1098 //consider & use space after each word Now draw a Huffman tree and generate codes.
- The size value of various nodes in a weight balanced tree areleaf – zerointernal node – size of it’s two childrenis this true?a) trueb) falseIn a tree, a(n) ____ is a node that has at least one child. Question 6 options: branch interior node root leafPROLOG PROGRAMMING LANGUAUGE PLEASE (* First argument is a BST and the second argument is a file name. Visits the tree nodes in preorder recursively and writes its data to the file separated by spaces. *) preOrderWrite/2
- Java / Trees: *Please refer to attached image* What is depth of Node W in this tree? Multiple chocie. 0 1 2 3 4 10public int insert(int value);/* Creates a node with the parameter as its value* and inserts the node into the tree in the appropriate position.* This method should call upon the recursive insert_r() helper method* you write.** we will make the assumption for this assignment (and implement it so) * that the values in a BST are unique... no duplicates** If we insert a value and it succeeded... adds it to the tree* return the element that was added** If we try to add an element and it is a duplicate, return the element* but make no changes to the tree... no new nodes added** @param int value to be inserted in tree* @return integer that was insrted after insertion is done.** Example 1: Suppose we have the tree below:** (27)* / \* (4) (29)* / \* (1) (8)* After calling insert(5), we'd obtain the following tree:** (27)* / \* (4) (29)* / \* (1) (8)* /* (5)** Example 2: Suppose we have the tree below:** (87)* / \* (42) (128)* \ \* (46) (145)* /* (44)** After calling insert(147), we'd obtain the…Complete the findMin() method. This method should return the minimum of the values in the tree. If findMin was called with the tree pictured before, it should return 1. class Main { public static void main(String[] args) { BinaryTree.main(args); } } public class BinaryTree { public BinaryTree() { root = null; } public BinaryTree(Node node) { root = node; } static class Node { Node(int d, Node l, Node r) { data = d; left = l; right = r; } public int data; public Node left; public Node right; } public Node root; public int findMin() { return -1; } public static void main(String[] args) { BinaryTree myTree = new BinaryTree(); myTree.root = new Node(10, new Node(5, new Node(20, null, null), new Node(25, new Node(15, new Node(30, null, null), new Node(35, null, null)), null)), new Node(40, null, null)); System.out.println(myTree.findMin()); } }