Create a function that would accept and integer as level/depth of the tree and display the tree-like output. • use this function definition: function tree(num) {} example code and output: tree(4) 1 101 111 11011 another example code and output: tree(8) 1 101 111 11011 11111 1110111 1111111 111101111
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Implement the following JavaScript function
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- Computer Science Exercise: shape [★★★] Write a function same_shape : 'a tree -> 'b tree -> bool that determines whether two trees have the same shape, regardless of whether the values they carry at each node are the same. Hint: use a pattern match with three branches, where the expression being matched is a pair of trees. please use Ocaml for the codingData Structures and Algorithms in C/C++ implement a variety of functions that operate on binary trees (the binary tree implementation from the book). You will be asked to test these functions on the following two trees (element data type is int): implement a variety of functions that operate on binary trees (the binary tree implementation from the book). You will be asked to test these functions on the following two trees (element data type is int): a)int count_leaves(BiTree *tree); Returns the number of leaf nodes in the tree. b) int count_non_leaves(BiTree *tree); Returns the number of non-leaf nodes in the tree. c) int get_height(BiTree *tree); Returns the height of the tree. d) void print_pre_order(BiTree *tree, void (*print)(const void *data)) Prints the elements of the tree to stdout using a pre-order traversal. The print parameter should contain the logic to print the data held in each node in the tree. e) void print_in_order(BiTree *tree, void (*print)(const void…Given the following struct that represents a binary tree: struct Node { int key: Node "parent; Node "left; Node "right; Nodelint k) : key(k), parent(nullptr), left(nullptr), right(nullptr) (I: 1: Write a recursive function that prints out the nodes in a tree stored using the above structure in order to cout. The function prints the depth (root depth is at 0) and key of that node separated by a colon (Example "O: 10\n" for root with key 10). Your function CAN NOT create any local variables and can only use what is passed to the function. Use the below function signature (NOTE: this is not a class method). void inorderAndDepth(Node "node, int depth)
- please convert the code to C language Source Code: // CPP program for above approach#include <bits/stdc++.h>using namespace std; // A Binary Tree Nodestruct node{ struct node *left, *right; int data;}; // Utility function to// create a new tree nodenode* newNode(int key){node *temp = new node;temp->data= key;temp->left = temp->right = NULL;return temp;} // Function to print all// non-root nodes that// don't have a siblingvoid printSingles(struct node *root){ // Base caseif (root == NULL)return; queue<node *> q1;q1.push(root); int flag=0;vector<int> v; // While q1 is not emptywhile(q1.empty() == false){ struct node * temp=q1.front(); q1.pop(); // Check if temp->left is not // NULL and temp->right is NULL if(temp->left != NULL && temp->right == NULL) { flag=1; v.push_back(temp->left->data); } // Check if temp->left is equal // NULL and temp->right is not NULL…struct insert_into_bst { // Function takes a constant Book as a parameter, inserts that book indexed by // the book's ISBN into a binary search tree, and returns nothing. void operator()(const Book& book) { // // TO-DO (7) ||| ///// // Write the lines of code to insert the key (book's ISBN) and value // ("book") pair into "my_bst". END-TO-DO (7) | } std::map& my_bst; };Create a function that takes an array that represent a Binary Tree and a value and return true if the value is in the tree and, false otherwise. Examples valueInTree (arr1, 5) valueInTree (arr1, 9) valueInTree (arr2, 51) - true 1 false false
- Write a function same_shape : 'a tree -> 'b tree -> bool that determines whether two trees have the same shape, regardless of whether the values they carry at each node are the same. Hint: use a pattern match with three branches, where the expression being matched is a pair of trees.Code in C++ only In a rooted tree, the lowest common ancestor (or LCA for short) of two vertices u and v is defined as the lowest vertex that is ancestor of both that two vertices. Given a tree of N vertices, you need to answer the question of the form "r u v" which means if the root of the tree is at r then what is LCA of u and v. Input: 12 23 14 2 142 242 Output: 1Programming questions:typedef struct node { int data; struct node *left, *right;}BT;The node structure of the binary tree (BT) is shown above. There is a binary tree T, please complete the function: int degreeone(BT *T) to compute how many degree 1 node in the BT. The T is the root pointer, and the function shoule return the total number of degree 1 node.
- Create a function that takes an array that represent a Binary Tree and a value and return true if the value is in the tree and, false otherwise. Examples valueInTree (arr1, 5) true valueInTree (arr1, 9) → false valueInTree (arr2, 51) → falseAnalyze the below code carefully and figure out what each fragment is doing.i)public void func (Tree root){func(root.left);func(root.right);cout<< root.data;}ii)public void func(Tree root){cout <<root.data;func(root.left);func(root.right);}Create a binary linked tree, and traverse the tree by using the recursive function. The structure of the tree is as follow: You should input the nodes in pre-order sequence. If a child of a node is NULL, input a space. Write the function of create binary tree, pre-order to print the nodes, in-order to print the nodes and post-order to print the nodes. Count the height of the tree. Header file typedef char ElemType; typedef struct node//define the type of binary tree node { }BTnode; Source file #include <stdio.h> #include <stdlib.h> #include "tree.h" BTnode * createTree()//create the binary tree,return the root { BTnode *tnode;// tnode is the root char elem; ;//input the character //if the input is a space,set the pointer as NULL Else// if the input is not a space,generate the binary node and create its left sub-tree and right…