Print binary heap (Java implementation)

Print binary heap: Using Hierarchical relationships

I am here to sort the heap first and then execute the print heap in sort printastree ()

public class Maxheap<t extends Comparable<?    Super t>> {private t[] data;    private int size;    private int capacity;        public maxheap (int capacity) {this.capacity = capacity;        this.size = 0;    This.data = (t[]) new Comparable[capacity + 1];        } public Maxheap (t[] arr) {//heapify, number of build heaps capacity = arr.length;        data = (t[]) new Comparable[capacity + 1];        System.arraycopy (arr, 0, data, 1, arr.length);        size = Arr.length;        for (int i = SIZE/2; I >= 1; i--) {Shiftdown (i);    }} public int size () {return this.size;    } public int getcapacity () {return this.capacity;    } public boolean IsEmpty () {return size = = 0;    } public T Seekmax () {return data[1];            } public void Swap (int i, int j) {if (i! = j) {T temp = data[i];            Data[i] = Data[j];        DATA[J] = temp; }} public void Insert (T item) {size++;        Data[size] = Item;    Shiftup (size);        } public T Popmax () {swap (1, size--);        Shiftdown (1);    return data[size + 1];  } public void Shiftup (int.) {while (Child > 1 && data[child].compareto (DATA[CHILD/2]) > 0)            {Swap (child, CHILD/2);        Child/= 2;  }}/** * @param the lower corner of an element in the data array * @param the lower corner of an element in the B data array * @return which element is large returns the lower corner of the label */private int max (int a, int b) {if (Data[a].compareto (Data[b]) < 0) {//if DATA[B] Big return b;//returns B} el SE {//if DATA[A] big return a;//Returns a}}/** * @param the lower corner of an element in the data array * @param the element in the B data array Subscript * @param the lower corner of an element in the C data array * @return which element returns the lower corner of the index * */private int max (int A, int b, int c) {in        T biggest = Max (A, b);        Biggest = Max (biggest, c);    Return biggest; } public void Shiftdown (int father) {while (true) {intLchild = Father * 2;            int rchild = Father * 2 + 1;                int newfather = father;//Here does not matter, if you change the following return to break, it must be assigned the IF (Lchild > Size) {//If there is no left and right child            Return            } else if (Rchild > Size) {//If there is no right child Newfather = max (father, Lchild);            } else {//If there are left and right children Newfather = max (father, Lchild, rchild);            if (Newfather = = father) {//If the original parent node is the largest of the three, then do not continue to tidy up the heap return;                The else {//parent node is not the largest, then swap the big child up, then continue the heap adjustment until the big root heap is satisfied (Newfather, father); Father = newfather;//is equivalent to continuing Shiftdown (Newfather). If Newfather turned out to be father's left child, it would be equivalent to Shiftdown (2*father)}}} public static <t extends comparable< ?        Super t>> void Sort (t[] arr) {int len = arr.length;        maxheap<t> maxheap = new maxheap<> (arr);        Maxheap.printastree (); for (int i = len-1; I >= 0; i--) {Arr[i] = maxHeap.popmax ();            }} public static void Printarr (object[] arr) {for (Object O:arr) {System.out.print (o);        System.out.print ("\ t");    } System.out.println (); The public void printspace (int n) {//Prints n spaces (where ' \ t ' is substituted) for (int i = 0; i < n; i++) {SYSTEM.OUT.PR        intf ("%3s", "" "); }} public void Printastree () {int linenum = 1;//First Traverse first line int lines = (int) (Math.log (size)/Math.log        (2) + 1;//lines is the number of layers of the heap int spacenum = (int) (Math.pow (2, lines)-1); for (int i = 1; i <= size;) {//because in [1...size] left closed right closed interval data, data[0] no data//each layer is printed this interval [2^ (Layer-1) ... (2^ layers)-1]. If the number in the heap is not enough (2^ layers)-1, then print to size.            So take min ((2^ layer) -1,size).                for (int j = (int) Math.pow (2, LineNum-1), J <= math.min (size, (int) Math.pow (2, LineNum)-1); J + +) { Printspace (Spacenum); Print Spacenum space System.out.printf ("%3s", Data[j]);//print Data System.out.printf ("%3s", "" ");//Picture green Box Printspace (spacenum);//print spacenum a space i++;//1 per element printed            } linenum++;            Spacenum = SPACENUM/2;        System.out.println ();        }} public static void Main (String args[]) {integer[] arr = {3, 5, 1, 7, 2, 9, 8, 0, 4, 6, 1, 3, 6, 1, 1};    Sort (arr); }}

Execution Result:

Print binary heap (Java implementation)