Question Answered step-by-step [Model: lessen organic product mango 50 will diminish the absolute… [Model: lessen organic product mango 50 will diminish the absolute mango to 50 from100.]o [Another Example: decrease goldfish 50 will erase the hub as goldfish isdecreased to 0]• erase: this order erases a thing from a given tree.o [Example: erase natural product avocado will erase the avocado hub from thenatural product tree]• delete_name: this order delte the whole tree of a given name.o [Example: delete_name creature will erase the creature tree as well ascreature hub from the name tree]Test Input:4 28 21fishcreaturebirdorganic productcreature feline 30fish goldfish 50creature canine 20bird blackbird 10creature bear 10organic product mango 100creature gator 50creature tiger 3creature lion 3fish swordfish 10creature deer 5creature cow 15fish garfish 5fish catfish 55fish salmon 40bird crow 20bird dove 10bird flamingo 15organic product apple 50organic product banana 50organic product nectarine 10organic product coconut 10organic product peach 40organic product apricot 30organic product avocado 25organic product cherry 100organic product cranberry 100creature horse 6search organic product avocadosearch fish tilapiasearch creature cowsearch bird crowsearch bird cowsearch creature felineitem_before creature deerheight_balance creatureheight_balance birdheight_balance fishsearch bloom rosecount creaturecount organic producterase creature felinesearch creature felinecount creatureerase fish swordfisherase organic product avocadodelete_name creaturelessen natural product mango 50search organic product mangoYield Specification: all the result to an out.txt record as well with respect to the control center. You arepermitted to utilize a worldwide variable for outfile pointer to improve on your capacityboundaries.In the wake of building the tree, you ought to print the trees in inorder in the predetermined organizationdisplayed in the example yield roar.Test Output:creature bird fish organic product===animal===crocodile bear feline cow deer canine pony lion tiger===bird===blackbird crow dove flamingo===fish===catfish garfish goldfish salmon swordfish===fruit===apple apricot avocado banana cherry coconut cranberrymango nectarine peach25 avocado found in organic producttilapia not found in fish15 cow saw as in creature20 crow found in birdcow not found in bird30 feline viewed as in creaturething before deer: 4creature: left stature 1, right tallness 3, contrast 2, not adjustedbird: left tallness – 1, right stature 2, contrast 3, not adjustedfish: left stature 1, right tallness 1, distinction 0, adjustedThere is no such thing as bloomcreature count 142organic product count 515feline erased from creaturefeline not viewed as in creaturecreature count 112swordfish erased from fishavocado erased from natural productcreature erasedmango decreased50 mango found in natural productExecution Restriction.:1. You need to utilize the accompanying design. You are permitted to change the construction if necessary.2. typedef struct itemNode{scorch name[MAXLEN];int count;struct itemNode *left, *right;}itemNode;typedef struct treeNameNode{scorch treeName[MAXLEN];struct treeNameNode *left, *right;itemNode *theTree;}treeNameNode;2.In expansion to run of the mill elements of tree execution, you should haveto carry out the accompanying capacities:I. createTreeNameNode()ii. treeNameNode* buildNameTree(…..): Based on the information in the record, it willembed them to the name tree and afterward at long last return the foundation of thename treeiii. traverse_in_traverse(treeNameNode *root): this capacity takes the foundation of thename tree and prints the information of the name tree and the relating thing trees in the organizationdisplayed in the example yield. You can call other capacity from this capacity on a case by case basis.iv. treeNameNode * searchNameNode(treeNameNode * root, singe treeName[50]):This capacity takes a name string and search this name in the name tree and returns that hub.This capacity will help you a great deal to go to the thing tree.v. Note that you could have to box separate inclusion and different capacities for name tree and thingtree. [I mean just two inclusion function.]vi. Every one of the numbers and result should be delivered from the trees.#incorporate #incorporate #incorporate #characterize MAXLEN 30Document *outFile;typedef struct itemNode {singe name[MAXLEN];int count;struct itemNode *left, *right;} itemNode;typedef struct treeNameNode {singe treeName[MAXLEN];struct treeNameNode *left, *right;itemNode *theTree;} treeNameNode;treeNameNode *createTreeNameNode(char *name) {treeNameNode *nameNode = (treeNameNode *)malloc(sizeof(treeNameNode));strcpy(nameNode->treeName, name);nameNode->left = NULL;nameNode->right = NULL;nameNode->theTree = NULL;return nameNode;}itemNode *createItemNode(char *name, int count) {itemNode *node = (itemNode *)malloc(sizeof(itemNode));strcpy(node->name, name);hub >count = count;hub >left = NULL;hub >right = NULL;bring hub back;}treeNameNode *insertNameNode(treeNameNode *root, treeNameNode *newNode) {on the off chance that (root == NULL) {  return newNode;} else {  // root name > newNode name  if (strcmp(root->treeName, newNode->treeName) > 0) {    // on the off chance that a component exists to one side of ‘root’ call once more    if (root->left != NULL) {      root->left = insertNameNode(root->left, newNode);    } else {      root->left = newNode;    }  }  // root name < newNode name OR they are equivalent strings  else {    // on the off chance that a component exists to one side of 'root' call once more    on the off chance that (root->right != NULL) {      root->right = insertNameNode(root->right, newNode);    } else {      root->right = newNode;    }  }}bring root back;}// determination from insertNameNode()itemNode *insertItemNode(itemNode *treeRoot, itemNode *newNode) {on the off chance that (treeRoot == NULL) {  return newNode;} else {  // treeRoot name > newNode name  if (strcmp(treeRoot->name, newNode->name) > 0) {    // on the off chance that a component exists to one side of it’s ‘root’ call once more    on the off chance that (treeRoot->left != NULL) {      treeRoot->left = insertItemNode(treeRoot->left, newNode);    } else {      treeRoot->left = newNode;    }  }  // treeRoot name < newNode name OR they are equivalent strings  else {    // on the off chance that a component exists to one side of it's 'root' call once more    on the off chance that (treeRoot->right != NULL) {      treeRoot->right = insertItemNode(treeRoot->right, newNode);    } else {      treeRoot->right = newNode;    }  }}return treeRoot;}treeNameNode *buildNameTree(FILE *inFile, int N) {if (inFile == NULL) {  bring NULL back;}treeNameNode *nameRoot = NULL;roast name[MAXLEN];for (int x = 0; x < N; x++) {  fscanf(inFile, "%s", name);  treeNameNode *newNameNode = createTreeNameNode(name);  nameRoot = insertNameNode(nameRoot, newNameNode);}return nameRoot;}treeNameNode *searchNameNode(treeNameNode *root, roast treeName[50]) {if (root == NULL) {  bring NULL back;}treeNameNode *nameNode;if (strcmp(root->treeName, treeName) == 0) {  bring root back;} else if (strcmp(root->treeName, treeName) > 0) {  nameNode = searchNameNode(root->left, treeName);} else {  nameNode = searchNameNode(root->right, treeName);}return nameNode;Image transcription textA new computer does a calculations inb hours and an old computer does ccalculations in d minutes I… Show more… Show moreImage transcription textThe Five Computing Disciplines 1.Match each discipline to its descriptionDiscipline Description Invo… Show more… Show more  Computer Science Engineering & Technology Java Programming COMPUTER S 113 Share QuestionEmailCopy link Comments (0)