平衡树，遵循了二叉排序树的原则，小的往左插，也可以认为是最优二叉排序树，因为它在插入的过程中通过每个节点的平衡因子判断自己是是否平衡，不平衡就会旋转树，使值再次遵循平衡树规则
- 这里直接上代码了：
/**
* Description :平衡树
*
* Author:yang
*
* Email:[email protected]
*
* Date: 2018/12/19
*/
public class AVLTree> {
TreeNode root; //根节点
int size = 0;
//平衡因子
private static final int LH = 1; //左子树大
private static final int RH = -1; //右子树大
private static final int EH = 0; //左右一样大
/**
* 节点
*
* @param
*/
public class TreeNode> {
Y item; //值
int balance = 0; //平衡因子
TreeNode left; //左子树
TreeNode right; //右子树
TreeNode parent; //父节点
private TreeNode(Y item, TreeNode parent) {
this.item = item;
this.parent = parent;
this.balance = 0;
this.left = null;
this.right = null;
}
}
/**
* 插入数据
*
* @param item
*/
public boolean addAVLTree(Y item) {
TreeNode t = root;
//根节点为空就直接赋值
if (t == null) {
root = new TreeNode<>(item, null);
root.balance = 0;
size = 1;
return true;
} else {
TreeNode parent = null;
//先找到要插入的位置
while (t != null) {
parent = t;
//item比t的位置小
if (item.compareTo(t.item) < 0) {
t = t.left;
} else if (item.compareTo(t.item) > 0) {
t = t.right;
} else { //相等就直接返回false
return false;
}
}
//现在parent就是你要插入的位置
TreeNode treeNode = new TreeNode<>(item, parent);
if (item.compareTo(parent.item) < 0) {
parent.left = treeNode;
} else {
parent.right = treeNode;
}
//节点已经放到了树上 现在往上修复平衡因子
while (parent != null) {
if (item.compareTo(parent.item) < 0) {
parent.balance++;
} else {
parent.balance--;
}
//如果平衡因子是0 则说明添加的树正好把位置补正了 直接退出
if (parent.balance == 0) {
break;
}
//如果修复过后因子等于2或-2了 就去转树
if (parent.balance == 2) {
leftBalance(parent);
break;
} else if (parent.balance == -2) {
rightBalance(parent);
break;
} else { //继续往上走
parent = parent.parent;
}
}
}
size++;
return true;
}
/**
* 查询所有树
*
* @param root
*/
public void showAllAVLTree(TreeNode root) {
LinkedList> list = new LinkedList<>();
//入队
list.offer(root);
while (!list.isEmpty()) {
//直接出队 输出
TreeNode node = list.pop();
System.out.print(node.item + " ");
if (node.left != null) {
list.offer(node.left);
}
if (node.right != null) {
list.offer(node.right);
}
}
}
/**
* 左边的平衡因子修复
*
* @param t
*/
private void leftBalance(TreeNode t) {
//先获取要转的节点的左孩子 根据它的因子判断是直接右转 还是先左转再右转
TreeNode tl = t.left;
switch (tl.balance) {
case LH: //直接右转
rightRotate(t);
t.balance = EH;
tl.balance = EH;
break;
case RH: //左转再右转
//先获取tl的右孩子
TreeNode tlr = tl.right;
//再根据tlr的平衡因子判断他们要修改后的平衡因子
switch (tlr.balance) {
case LH:
t.balance = RH;
tl.balance = EH;
tlr.balance = EH;
break;
case RH:
t.balance = EH;
tl.balance = LH;
tlr.balance = EH;
break;
case EH:
t.balance = EH;
tl.balance = EH;
tlr.balance = EH;
break;
default:
break;
}
//先左转t的左孩子 在右转t
leftRotate(t.left);
rightRotate(t);
break;
}
}
/**
* 右边的平衡因子修复
*
* @param t
*/
private void rightBalance(TreeNode t) {
TreeNode tr = t.right;
switch (tr.balance) {
case RH:
leftRotate(t);
t.balance = EH;
tr.balance = EH;
break;
case LH:
TreeNode trl = tr.left;
switch (trl.balance) {
case RH:
t.balance = LH;
tr.balance = EH;
trl.balance = EH;
break;
case LH:
t.balance = EH;
tr.balance = RH;
trl.balance = EH;
break;
case EH:
t.balance = EH;
tr.balance = EH;
trl.balance = EH;
break;
default:
break;
}
rightRotate(t.right);
leftRotate(t);
break;
}
}
/**
* 左旋 分为下面三种情况
* 注：这里其实不要管y的右子树 只要管他有没有左子树就行了
*
* / x x x
* / / \ / \ / \
* / a y a y a y
* / / \ / \
* / b c b c
*
* @param x
*/
private void leftRotate(TreeNode x) {
if (x != null) {
//先取到y
TreeNode y = x.right;
//把x的右孩子指向y的左孩子
x.right = y.left;
//如果这里y有左孩子 就把y的左孩子父节点赋值
if (y.left != null) {
y.left.parent = x;
}
//再把y的父节点赋为x
y.parent = x.parent;
//如果x的父节点是null 就说明x是根节点
if (x.parent == null) {
root = y;
} else {
//这里判断x是父节点的左孩子还是右孩子
if (x.parent.left == x) {
x.parent.left = y;
} else if (x.parent.right == x) {
x.parent.right = y;
}
}
//x作为y左孩子
y.left = x;
x.parent = y;
}
}
/**
* 右旋转
*
* / y y y
* / / \ / \ / \
* / z a z a z a
* / / \ / \
* / b c b c
*
* @param y
*/
private void rightRotate(TreeNode y) {
if (y != null) {
TreeNode z = y.left;
//step1
y.left = z.right;
if (z.right != null) {
z.right.parent = y;
}
//step2
z.parent = y.parent;
if (y.parent == null) {
root = z;
} else {
if (y.parent.left == y) {
y.parent.left = z;
} else if (y.parent.right == y) {
y.parent.right = z;
}
}
//step3
z.right = y;
y.parent = z;
}
}
}