Recently in the study of data structure on the balance of binary tree knowledge, look at strict teacher's thinking, feel a bit difficult to write a recursive construction method in Java, because the recursion needs to pass in the reference, so the feeling is to achieve a more troublesome, so the first thought of using a non-recursive way to achieve the construction of balanced binary tree.

Using a non-recursive approach, the idea is simple, is to define a balance factor for each node of the attribute, when the success of inserting a data into the tree, I will go back to see if there is no balance factor of the absolute value equals 2 of the node, if there is, it needs to be rotated. At that time, the idea was limited to these, the contact with Java not long, visual inspection if implemented, a bit difficult.

Therefore, the Internet query a blog to write code, although do not know who the original author, but still affixed to my reference blog address: http://blog.csdn.net/zxman660/article/details/7940190

Among them, I think there is a problem is that I define the node using generics, that is, the node value is also used in the generic <e>, so in comparison size, I do not know, may be I contact Java for not too long reason, when I refer to the above blog's ideas, incredibly this realization.

void Compare (e element, node node) {comparable<? super e> E = (comparable<? super e>) element; int cmp = E.comp Areto (node.element); if (CMP > 0)//greater than else if (CMP = = 0)//equals else//less}

Through this study, although contact with the balance of binary tree construction, but still feel learned a lot of knowledge, and the use of Java has more understanding.

With the above sentiment, feel or write a blog to collect, and so you have time, and then add the node deletion code.

Perhaps I code in the comments on the idea of a limited, if there is not understand the place, you can refer to the above blog address content.

Package util;/** * Balanced binary tree * Definition: First it is a special two-fork sorting tree, followed by its Saozi right subtree is balanced binary tree, * and Saozi right subtree depth difference not more than 1 * balance factor: can be defined as the depth of the Zuozi minus the depth of the right subtree * * The Balanced binary tree is the optimization of the two-fork sorting tree, which prevents the binary sort tree from finding the average time in the worst case of N, * binary sort tree in this time shape as a single linked list * Balanced binary tree The number of times to find elements does not exceed the depth of the tree, the time complexity is LOGN */public class Avltree<e > {/** root node */private node<e> root = number of elements in null;/** tree */private int size = 0;private static final int left_high = 1;PR ivate static final int right_high = -1;private static final int equal_high = 0;public Avltree () {}public Boolean Insertelem ENT (E element) {node<e> t = root;if (t = = null) {/** Copies the value to the root node whose parent is empty */root = new node (element, null); size = 1;return Tru e;} int cmp = 0; Node<e> parent; /** is used to save the parent node of the T *//** find the location */comparable< the node should be stored; Super e> E = (comparable<? super e>) Element;do{parent = t;cmp = E.compareto (t.element); if (CMP < 0) {t = T.left; }else if (cmp > 0) {t = t.right;} Else{return false;}} while (t! = null),/** stores the node in the appropriate location */node<e> child = new Node (element, parent), if (CMP < 0) {parent.left = child;} Else{parent.rigHT = child;} /** begins backtracking, modifying the BALABCE for the root node so that the corresponding adjustment */while (parent! = NULL) {CMP = E.compareto (parent.element); if (CMP < 0) { Parent.balance + +;} Else{parent.balance--;} if (parent.balance = = 0) {/** from the above nodes need not be modified, do not need to rotate the */break;} if (Math.Abs (parent.balance) = = 2) {fixafterinsertion (parent); break;} parent = parent.parent;} Size ++;return true;} private void Fixafterinsertion (Node<e> p) {if (p.balance = = 2) {leftbanance (P);} if (p.balance = =-2) {rightbalance (P);}} /** * Left balance operation, that is, the imbalance of the node T is because the left dial hand tree is too deep * * 1, if the new node is inserted into the left child of P-left subtree, then the right-hand operation can be directly * TLC */\ Right-spin/* LC RC-------------> LCL T */\//* LCL LCR LCLL LCR RC */* LCLL * 2, if a new node is inserted into the right subtree of P's left child, a sub-case discussion is required * Condition A: When P's left child's right child tree root node balance = Right_high * 11 4 */\/\/* 2 6 L 4 6 Right-handed 2 1 */\------->/\-------->//* 3 4 2 5 3 5 6 * \/* 5 3 * * * case B: When P's left child's right child tree root node balance = Left_high * 11 4 */\/\/* 2 6 left-handed 4 6 Right-handed 2 1 */\------->/-------->/\ * 3 4 2 3 5 6 *//* 5 3 5 * * Case C: When P's left child's right child root Balance of the node = equal_high * * 11 4 */\/\/* 2 7 L 4 7 right-hand 2 1 */ \------->/\-------->/\/* 3 4 2 6 3 5 6 7 */\/* 5 6 3 5 * */private void leftbanance (Node<e> t) {NODE<E&G T LC = T.left;switch (lc.balance) {Case left_high:/** new node inserted into T left child's left subtree, requires single right-hand processing */lc.balance = Equal_high;t.balance = Equal_ High;break;case right_high:/** new node is inserted into the left child's right subtree of T, requires a double spin process */node<e> rd = Lc.right;switch (rd.balance) {case Left_ HIGH:lc.balance = Equal_high;t.balanCe = right_high;break;case RIGHT_HIGH:lc.balance = left_high;t.balance = Equal_high;break;case Equal_high:t.balance = Equal_high;lc.balance = Equal_high;break;} Rd.balance = equal_high;/** Left-handed */left_rotate (T.left) to the left subtree of T,/** right-handed to T */right_rotate (t); break;}} /** * Right balance operation, that is, the imbalance of the node T is because the right subtree is too deep * * 1, if the new node is inserted into the right child right subtree of p, then the direct left-hand operation can be * * PR */\/* L R Left-hand operation P RR */ \----------->/\ * RL RRL RL RRR * * RRR * * * 2, if a new node is inserted into the left subtree of the right child of P, a sub-case discussion is required * case A: When P's right child's Zogen node balance = Left_high * *114 */\/\/* 2 3 right-hand 2 4 left-handed 1 3 */\-------> /\------->/\ * 4 5 6 326 5 */* 65 * * case B: When P's right child's Zogen node balance = Right_high * *114 */ \/\/* 2 3 right-handed 2 4 left-handed 1 3 */\-------> \------->//* 4 5 32 6 5 * \/* 66 5 * * * Case c: When P's right child's Zogen node balance = equal_high *114 */\/\/* 2 3 Right spin 2 4 left Spin 1 3 */\------->/\------->/\/* 4 5 6 326 7 5 */\/* 6 + 5 * */private void righ Tbalance (node<e> p) {node<e> rc = P.right;switch (rc.balance) {Case right_high:/** new node inserted into T right child's right subtree, requires single left-handed processing */rc.balance = Equal_high;p.balance = equal_high;break;case left_high:/** The new node is inserted into the left subtree of the right child of T, the double-spin processing is required for the */node<e> ld = Rc.left;switch (ld.balance) {case left_high:p.balance = Equal_high;rc.balance = Right_high;break;case RIGHT_HIGH: P.balance = Left_high;rc.balance = Equal_high;break;case equal_high:p.balance = equal_high;rc.balance = EQUAL_HIGH; break;} Ld.balance = equal_high;/** Right-handed */right_rotate (p.right) to the right subtree of P,/** to p for left-hand processing */left_rotate (p); break;}} /** * Left-handed operation * PR */\/* L R Left-hand operation P RR */\----------->/\ * RL RRL RL RR R * * RRR * */private void Left_rotate (node<e> p) {if (P! = null) {node<e> r = p.right;/** Get P's right subtree root node R*/p.ri Ght = r.left;/** The left subtree of R is transferred to the right subtree of P */if (r.left! = null) {/** If R's left subtree is not empty, the left sub-treeThe parent node of the tree is set to P*/r.left.parent = p;} R.parent = p.parent;/** modifies the parent node of R, modifies the parent node of P */if (p.parent = = null) {/** If the parent node of P is null, then r is the root node */root = r;} else if (p = = p.parent.left) {/** If P is the left child of its parent node, the left child of its parent node points to R*/p.parent.left = R;} else if (p = = p.parent.right) {/** If P is the right child of its parent node, point the right child of its parent node to R*/p.parent.right = R;} R.left = p;/** Set R's left child to P*/p.parent = r;/** the parent node of P is set to r*/}}/** * right-hand operation * PL */\ Right-hand operation/* L R-------------> ll p */\//* LL lrlll LR R */* LLL * */private void Right_rotate (node<e> p) {if (P! = NULL {node<e> L = p.left;/** gets P's Left child l*/p.left = l.right;/** turns the right subtree of L to the left subtree of P */if (l.right! = null) {/** If the right subtree of L is not empty, Set its parent node to p*/l.right.parent = p;} L.parent = p.parent;/** modifies the parent node of R to the parent of P */if (p.parent = = null) {/** If the parent node of P is null, that is, l is Root*/root = l;} else if (p = = p.parent.left) {/** If P is the left child of its parent node, the left child of P's parent node points to L*/p.parent.left = l;} else if (p = = p.parent.right) {/** If P is the right child of its parent node, the right child of P's parent node points to L*/p.parent.right = l;} L.right = p;/** Changes the right subtree of l to P*/p.parent = l;/** modifies the parent node of P to l*/}}/** the non-recursive way to traverse the balanced binary tree */public void Nrinordertraverse () {stack<node<e>> Stack = new stack<node<e>> (); Node<e> p = root;while (P! = NULL | |!stack.isempty ()) {while (P! = null) {Stack.push (p);p = P.left;} p = Stack.pop (); System.out.println (p.element);p = P.right;}} The node definition of/** balanced binary tree */class node<e>{e element;/** node balance factor */int balance = 0;/** left child node, right child node, parent */node<e> Ieft; Node<e> right; Node<e> parent;public node () {}public node (E element, node<e> parent) {this.element = Element;this.parent = Parent;} Public String toString () {return element + "bf=" + Balance;}} public static void Main (string[] args) {integer[] num = {5,8,2,0,1,-2,-9, 100}; avltree<integer> AVL = new avltree<integer> (); for (int i = 0; i < num.length; i++) {avl.insertelement (Num[i] );} Avl.nrinordertraverse ();}}