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本篇内容主要讲解“Java平衡二叉树怎么实现”，感兴趣的朋友不妨来看看。本文介绍的方法操作简单快捷，实用性强。下面就让小编来带大家学习“Java平衡二叉树怎么实现”吧!
简单来说，就是方便搜索的二叉树，是一种具备特定结构的二叉树，即，对于节点n，其左子树的所有节点的值都小于等于其值，其右子树的所有节点的值都大于等于其值。
以序列2,4,1,3,5,10,9,8为例，如果以二叉搜索树建树的方式，我们建立出来的逐个步骤应该为
第一步：
第二步：
第三步：
第四步：
第五步：
第六步：
第七步：
第八步：
按照不平衡的普通方法生成的二叉搜索树就是这么一个样子。其实现代码如下：
package com.chaojilaji.book.searchtree;
import com.chaojilaji.auto.autocode.utils.Json;
import java.util.Objects;
public class SearchTreeUtils {
public static SearchTree buildTree(SearchTree searchTree, Integer value) {
if (value >= searchTree.getValue()) {
if (Objects.isNull(searchTree.getRightChild())) {
SearchTree searchTree1 = new SearchTree();
searchTree1.setValue(value);
searchTree.setRightChild(searchTree1);
} else {
buildTree(searchTree.getRightChild(), value);
}
} else {
if (Objects.isNull(searchTree.getLeftChild())) {
SearchTree searchTree1 = new SearchTree();
searchTree1.setValue(value);
searchTree.setLeftChild(searchTree1);
} else {
buildTree(searchTree.getLeftChild(), value);
}
}
return searchTree;
}
public static void main(String[] args) {
int[] a = new int[]{2, 4, 1, 3, 5, 10, 9, 8};
SearchTree searchTree = new SearchTree();
searchTree.setValue(a[0]);
for (int i = 1; i < a.length; i++) {
searchTree = buildTree(searchTree,a[i]);
}
System.out.println(Json.toJson(searchTree));
}
}运行的结果如下：
{
"value": 2,
"left_child": {
"value": 1,
"left_child": null,
"right_child": null
},
"right_child": {
"value": 4,
"left_child": {
"value": 3,
"left_child": null,
"right_child": null
},
"right_child": {
"value": 5,
"left_child": null,
"right_child": {
"value": 10,
"left_child": {
"value": 9,
"left_child": {
"value": 8,
"left_child": null,
"right_child": null
},
"right_child": null
},
"right_child": null
}
}
}
}与我们的目标结果是一致的。
好了，那我们本节就完毕了。可是转过头可能你也发现了，直接生成的这个二叉搜索树似乎有点太长了，层数有点太多了，一般来说，一个长度为8的序列，四层结构的二叉树就可以表现出来了，这里却使用了六层，显然这样的结果不尽人意，同时太深的层数，也增加了查找的时间复杂度。
这就给我们的树提了要求，我们需要将目前构造出来的树平衡一下，让这棵二叉搜索树的左右子树“重量”最好差不多。
首先需要掌握两个概念
平衡因子
旋转
平衡因子就是对于这棵二叉搜索树的每个节点来说，其左子树的高度减去右子树的高度即为该节点的平衡因子，该数值能很快的辨别出该节点究竟是左子树高还是右子树高。在平衡二叉树中规定，当一个节点的平衡因子的绝对值大于等于2的时候，我们就认为该节点不平衡，需要进行调整。那么这种调整的手段称之为节点与节点的旋转，通俗来说，旋转就是指的节点间的指向关系发生变化，在c语言中就是指针指向的切换。
在调用旋转之前，我们需要判断整棵树是否平衡，即，这棵二叉搜索树的所有平衡因子是否有绝对值大于等于2的，如果有，就找出最小的一棵子树。可以确定的是，如果前一次二叉搜索树是平衡的，那么此时如果加一个节点进去，造成不平衡，那么节点从叶子开始回溯，找到的第一个大于等于2的节点势必为最小不平衡子树的根节点。
对于这棵最小不平衡的子树，我们需要得到两个值，即根节点的平衡因子a，以及左右子树根节点中平衡因子绝对值较大者的平衡因子b。
我们可以将需要旋转的类型抽象为以下四种：
1.左左型（正正型，即 a>0 && b>0）
左左型最后想要达到的目标是第二个节点成为根节点，第一个节点成为第二个节点的右节点。
所以用伪代码展示就是（设a1，a2，a3分别为图里面从上到下的三个节点）
a2的右子树 = (合并(a2的右子树,a1的右子树) + a1顶点值) 一起构成的二叉搜索树;
返回 a2
2.左右型（正负型，即 a>0 && b<0）
设a1，a2，a3分别为图里面从上到下的三个节点
首先应该通过将a3和a2调换上下位置，使之变成左左型，然后再调用左左型的方法就完成了。
从左右型调换成左左型，即将a2及其左子树成为a3左子树的一部分，然后将a1的左子树置为a3即可。
伪代码如下：
a3的左子树 = a2及其左子树与a3的左子树合并成的一棵二叉搜索树;
a1的左子树 = a3;
3.右右型（负负型，即 a<0 && b<0）
设a1，a2，a3分别为图里面从上到下的三个节点
右右型与左左型类似，要达到的目的就是a1成为a2左子树的一部分，伪代码为：
a2的左子树 = （合并a2的左子树和a1的左子树）+ a1顶点的值构成的二叉搜索树;
返回a2
4.右左型（负正型，即 a<0 && b>0）
设a1，a2，a3分别为图里面从上到下的三个节点
右左型需要先转换成右右型，然后在调用右右型的方法即可。
从右左型到右右型，即需要将a2及其右子树成为a3右子树的一部分，然后将a1的右子树置为a3即可。
伪代码如下：
a3的右子树 = a2及其右子树与a3的右子树合并成的一棵二叉搜索树;
a1的右子树 = a3;
从上面的分析可以得出，我们不仅仅需要实现旋转的方法，还需要实现合并二叉树等方法，这些方法都是基础方法，读者需要确保会快速写出来。
请读者朋友们根据上面的内容，先尝试写出集中平衡化的方法。
平衡二叉搜索树建树，需要在二叉搜索树建树的基础上加上平衡的过程，即子树之间指针转换的问题，同时，由于这种指针转换引起的子树的子树也会产生不平衡，所以上面提到的四种旋转调整方式都是递归的。
首先，先构建节点基础结构：
public class SearchTree {
private Integer value;
private SearchTree leftChild;
private SearchTree rightChild;
private Integer balanceNumber = 0;
private Integer height = 0;
}值，高度，平衡因子，左子树，右子树
这是计算二叉搜索树中每个平衡因子的基础，我们设最低层为高度1，则计算节点高度的代码为：
public static Integer countHeight(SearchTree searchTree) {
if (Objects.isNull(searchTree)) {
return 0;
}
searchTree.setHeight(Math.max(countHeight(searchTree.getLeftChild()),
countHeight(searchTree.getRightChild())) + 1);
return searchTree.getHeight();
}这里有个半动态规划的结论：当前节点的高度，等于左右子树的最大高度+1；这里的写法有点树形DP的味道。
public static void countBalanceNumber(SearchTree searchTree, MaxNumber max, SearchTree fatherTree, Integer type) {
if (Objects.nonNull(searchTree.getValue())) {
if (Objects.isNull(searchTree.getLeftChild())
&& Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(-searchTree.getRightChild().getHeight());
}
if (Objects.nonNull(searchTree.getLeftChild())
&& Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight());
}
if (Objects.isNull(searchTree.getLeftChild())
&& Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(0);
}
if (Objects.nonNull(searchTree.getLeftChild())
&& Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight()
- searchTree.getRightChild().getHeight());
}
}
if (Objects.nonNull(searchTree.getLeftChild())) {
countBalanceNumber(searchTree.getLeftChild(), max, searchTree, 1);
}
if (Objects.nonNull(searchTree.getRightChild())) {
countBalanceNumber(searchTree.getRightChild(), max, searchTree, 2);
}
}本质上讲，平衡因子就是左子树高度减去右子树高度，注意这里左右子树都有可能不存在，所以加入了一堆特判。
判断当前二叉树是否平衡
static class MaxNumber {
public Integer max;
public SearchTree childTree;
public SearchTree fatherTree;
public Integer flag = 0; // 0 代表自己就是根,1代表childTree是左子树，2代表childTree是右子树
}
public static MaxNumber checkBalance(SearchTree searchTree) {
MaxNumber max = new MaxNumber();
max.max = 0;
countBalanceNumber(searchTree, max, null, 0);
return max;
}
public static void countBalanceNumber(SearchTree searchTree, MaxNumber max, SearchTree fatherTree, Integer type) {
if (Objects.nonNull(searchTree.getValue())) {
if (Objects.isNull(searchTree.getLeftChild())
&& Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(-searchTree.getRightChild().getHeight());
}
if (Objects.nonNull(searchTree.getLeftChild())
&& Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight());
}
if (Objects.isNull(searchTree.getLeftChild())
&& Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(0);
}
if (Objects.nonNull(searchTree.getLeftChild())
&& Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight()
- searchTree.getRightChild().getHeight());
}
}
if (Objects.nonNull(searchTree.getLeftChild())) {
countBalanceNumber(searchTree.getLeftChild(), max, searchTree, 1);
}
if (Objects.nonNull(searchTree.getRightChild())) {
countBalanceNumber(searchTree.getRightChild(), max, searchTree, 2);
}
if (Math.abs(searchTree.getBalanceNumber()) >= Math.abs(max.max)) {
if (Math.abs(searchTree.getBalanceNumber()) == Math.abs(max.max)
&& max.childTree == null) {
max.childTree = searchTree;
max.fatherTree = fatherTree;
max.flag = type;
max.max = searchTree.getBalanceNumber();
}
if (Math.abs(searchTree.getBalanceNumber()) > Math.abs(max.max)) {
max.childTree = searchTree;
max.fatherTree = fatherTree;
max.flag = type;
max.max = searchTree.getBalanceNumber();
}
}
}其中，MaxNumber类是为了保存第一棵不平衡的子树而存在的结构，为了使这棵子树平衡之后能重新回到整棵树中，需要在MaxNumber中存储当前子树父节点，同时标明当前子树是父节点的左子树还是右子树，还是本身。
public static void getAllValue(SearchTree tree, Set<Integer> sets) {
if (Objects.isNull(tree)) return;
if (Objects.nonNull(tree.getValue())) {
sets.add(tree.getValue());
}
if (Objects.nonNull(tree.getLeftChild())) {
getAllValue(tree.getLeftChild(), sets);
}
if (Objects.nonNull(tree.getRightChild())) {
getAllValue(tree.getRightChild(), sets);
}
}
/**
* 合并两棵二叉搜索树
*
* @param a
* @param b
* @return
*/
public static SearchTree mergeTree(SearchTree a, SearchTree b) {
Set<Integer> vals = new HashSet<>();
getAllValue(b, vals);
for (Integer c : vals) {
a = buildTree(a, c);
}
return a;
}将一棵树转成数字集合，然后通过建树的方式建到另外一棵树上即可。
1.左左型旋转
/**
* 左左
*
* @param searchTree
* @return
*/
public static SearchTree leftRotate1(SearchTree father, SearchTree searchTree) {
SearchTree b = father;
SearchTree newRight = mergeTree(father.getRightChild(), searchTree.getRightChild());
newRight = buildTree(newRight, b.getValue());
countHeight(newRight);
while (Math.abs(checkBalance(newRight).childTree.getBalanceNumber()) >= 2) {
newRight = rotate(checkBalance(newRight).childTree);
countHeight(newRight);
}
searchTree.setRightChild(newRight);
return searchTree;
}2.右右型旋转
/**
* 右右
* @param father
* @param searchTree
* @return
*/
public static SearchTree rightRotate1(SearchTree father, SearchTree searchTree) {
SearchTree b = father;
SearchTree newLeft = mergeTree(father.getLeftChild(), searchTree.getLeftChild());
newLeft = buildTree(newLeft, b.getValue());
countHeight(newLeft);
while (Math.abs(checkBalance(newLeft).childTree.getBalanceNumber()) >= 2) {
newLeft = rotate(checkBalance(newLeft).childTree);
countHeight(newLeft);
}
searchTree.setLeftChild(newLeft);
return searchTree;
}3.左右型旋转
/**
* 左右
*
* @param searchTree
* @return
*/
public static SearchTree rightRotate2(SearchTree father, SearchTree searchTree) {
SearchTree a1 = father;
SearchTree a2 = searchTree;
SearchTree a3 = searchTree.getRightChild();
SearchTree newLeft = mergeTree(a2.getLeftChild(), a3.getLeftChild());
newLeft = buildTree(newLeft, a2.getValue());
countHeight(newLeft);
while (Math.abs(checkBalance(newLeft).childTree.getBalanceNumber()) >= 2) {
newLeft = rotate(checkBalance(newLeft).childTree);
countHeight(newLeft);
}
a3.setLeftChild(newLeft);
a1.setLeftChild(a3);
return a1;
}4.右左型旋转
/**
* 右左
*
* @param searchTree
* @return
*/
public static SearchTree leftRotate2(SearchTree father, SearchTree searchTree) {
SearchTree a1 = father;
SearchTree a2 = searchTree;
SearchTree a3 = searchTree.getLeftChild();
SearchTree newRight = mergeTree(a2.getRightChild(), a3.getRightChild());
newRight = buildTree(newRight, a2.getValue());
countHeight(newRight);
while (Math.abs(checkBalance(newRight).childTree.getBalanceNumber()) >= 2) {
newRight = rotate(checkBalance(newRight).childTree);
countHeight(newRight);
}
a3.setRightChild(newRight);
a1.setRightChild(a3);
return a1;
}旋转调用函数：
public static SearchTree rotate(SearchTree searchTree) {
int a = searchTree.getBalanceNumber();
if (Math.abs(a) < 2) {
return searchTree;
}
int b = Objects.isNull(searchTree.getLeftChild()) ? 0
: searchTree.getLeftChild().getBalanceNumber();
int c = Objects.isNull(searchTree.getRightChild()) ? 0
: searchTree.getRightChild().getBalanceNumber();
if (a > 0) {
if (b > 0) {
// TODO: 2022/1/13 左左
searchTree = leftRotate1(searchTree, searchTree.getLeftChild());
} else {
// TODO: 2022/1/13 左右
searchTree = rightRotate2(searchTree, searchTree.getLeftChild());
searchTree = leftRotate1(searchTree, searchTree.getLeftChild());
}
} else {
if (c > 0) {
// TODO: 2022/1/13 右左
searchTree = leftRotate2(searchTree, searchTree.getRightChild());
searchTree = rightRotate1(searchTree, searchTree.getRightChild());
} else {
// TODO: 2022/1/13 右右
searchTree = rightRotate1(searchTree, searchTree.getRightChild());
}
}
return searchTree;
}package com.chaojilaji.book.searchtree;
import com.chaojilaji.auto.autocode.utils.Json;
import com.chaojilaji.book.tree.Handle;
import com.chaojilaji.book.tree.Tree;
import org.omg.CORBA.OBJ_ADAPTER;
import java.util.HashSet;
import java.util.Objects;
import java.util.Set;
public class SearchTreeUtils {
static class MaxNumber {
public Integer max;
public SearchTree childTree;
public SearchTree fatherTree;
public Integer flag = 0; // 0 代表自己就是根,1代表childTree是左子树，2代表childTree是右子树
}
public static SearchTree rotate(SearchTree searchTree) {
int a = searchTree.getBalanceNumber();
if (Math.abs(a) < 2) {
return searchTree;
}
int b = Objects.isNull(searchTree.getLeftChild()) ? 0 : searchTree.getLeftChild().getBalanceNumber();
int c = Objects.isNull(searchTree.getRightChild()) ? 0 : searchTree.getRightChild().getBalanceNumber();
if (a > 0) {
if (b > 0) {
// TODO: 2022/1/13 左左
searchTree = leftRotate1(searchTree, searchTree.getLeftChild());
} else {
// TODO: 2022/1/13 左右
searchTree = rightRotate2(searchTree, searchTree.getLeftChild());
searchTree = leftRotate1(searchTree, searchTree.getLeftChild());
}
} else {
if (c > 0) {
// TODO: 2022/1/13 右左
searchTree = leftRotate2(searchTree, searchTree.getRightChild());
searchTree = rightRotate1(searchTree, searchTree.getRightChild());
} else {
// TODO: 2022/1/13 右右
searchTree = rightRotate1(searchTree, searchTree.getRightChild());
}
}
return searchTree;
}
public static void getAllValue(SearchTree tree, Set<Integer> sets) {
if (Objects.isNull(tree)) return;
if (Objects.nonNull(tree.getValue())) {
sets.add(tree.getValue());
}
if (Objects.nonNull(tree.getLeftChild())) {
getAllValue(tree.getLeftChild(), sets);
}
if (Objects.nonNull(tree.getRightChild())) {
getAllValue(tree.getRightChild(), sets);
}
}
/**
* 合并两棵二叉搜索树
*
* @param a
* @param b
* @return
*/
public static SearchTree mergeTree(SearchTree a, SearchTree b) {
Set<Integer> vals = new HashSet<>();
getAllValue(b, vals);
for (Integer c : vals) {
a = buildTree(a, c);
}
return a;
}
/**
* 左左
*
* @param searchTree
* @return
*/
public static SearchTree leftRotate1(SearchTree father, SearchTree searchTree) {
SearchTree b = father;
SearchTree newRight = mergeTree(father.getRightChild(), searchTree.getRightChild());
newRight = buildTree(newRight, b.getValue());
countHeight(newRight);
while (Math.abs(checkBalance(newRight).childTree.getBalanceNumber()) >= 2) {
newRight = rotate(checkBalance(newRight).childTree);
countHeight(newRight);
}
searchTree.setRightChild(newRight);
return searchTree;
}
/**
* 右左
*
* @param searchTree
* @return
*/
public static SearchTree leftRotate2(SearchTree father, SearchTree searchTree) {
SearchTree a1 = father;
SearchTree a2 = searchTree;
SearchTree a3 = searchTree.getLeftChild();
SearchTree newRight = mergeTree(a2.getRightChild(), a3.getRightChild());
newRight = buildTree(newRight, a2.getValue());
countHeight(newRight);
while (Math.abs(checkBalance(newRight).childTree.getBalanceNumber()) >= 2) {
newRight = rotate(checkBalance(newRight).childTree);
countHeight(newRight);
// System.out.println(Json.toJson(newRight));
}
a3.setRightChild(newRight);
a1.setRightChild(a3);
return a1;
}
/**
* 右右
* @param father
* @param searchTree
* @return
*/
public static SearchTree rightRotate1(SearchTree father, SearchTree searchTree) {
SearchTree b = father;
SearchTree newLeft = mergeTree(father.getLeftChild(), searchTree.getLeftChild());
newLeft = buildTree(newLeft, b.getValue());
countHeight(newLeft);
// TODO: 2022/1/13 合并后的也有可能有问题
while (Math.abs(checkBalance(newLeft).childTree.getBalanceNumber()) >= 2) {
newLeft = rotate(checkBalance(newLeft).childTree);
countHeight(newLeft);
// System.out.println(Json.toJson(newLeft));
}
searchTree.setLeftChild(newLeft);
return searchTree;
}
/**
* 左右
*
* @param searchTree
* @return
*/
public static SearchTree rightRotate2(SearchTree father, SearchTree searchTree) {
SearchTree a1 = father;
SearchTree a2 = searchTree;
SearchTree a3 = searchTree.getRightChild();
SearchTree newLeft = mergeTree(a2.getLeftChild(), a3.getLeftChild());
newLeft = buildTree(newLeft, a2.getValue());
countHeight(newLeft);
while (Math.abs(checkBalance(newLeft).childTree.getBalanceNumber()) >= 2) {
newLeft = rotate(checkBalance(newLeft).childTree);
countHeight(newLeft);
}
a3.setLeftChild(newLeft);
a1.setLeftChild(a3);
return a1;
}
public static MaxNumber checkBalance(SearchTree searchTree) {
MaxNumber max = new MaxNumber();
max.max = 0;
countBalanceNumber(searchTree, max, null, 0);
return max;
}
public static Integer countHeight(SearchTree searchTree) {
if (Objects.isNull(searchTree)) {
return 0;
}
searchTree.setHeight(Math.max(countHeight(searchTree.getLeftChild()), countHeight(searchTree.getRightChild())) + 1);
return searchTree.getHeight();
}
public static void countBalanceNumber(SearchTree searchTree, MaxNumber max, SearchTree fatherTree, Integer type) {
if (Objects.nonNull(searchTree.getValue())) {
if (Objects.isNull(searchTree.getLeftChild()) && Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(-searchTree.getRightChild().getHeight());
}
if (Objects.nonNull(searchTree.getLeftChild()) && Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight());
}
if (Objects.isNull(searchTree.getLeftChild()) && Objects.isNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(0);
}
if (Objects.nonNull(searchTree.getLeftChild()) && Objects.nonNull(searchTree.getRightChild())) {
searchTree.setBalanceNumber(searchTree.getLeftChild().getHeight() - searchTree.getRightChild().getHeight());
}
}
if (Objects.nonNull(searchTree.getLeftChild())) {
countBalanceNumber(searchTree.getLeftChild(), max, searchTree, 1);
}
if (Objects.nonNull(searchTree.getRightChild())) {
countBalanceNumber(searchTree.getRightChild(), max, searchTree, 2);
}
if (Math.abs(searchTree.getBalanceNumber()) >= Math.abs(max.max)) {
if (Math.abs(searchTree.getBalanceNumber()) == Math.abs(max.max) && max.childTree == null) {
max.childTree = searchTree;
max.fatherTree = fatherTree;
max.flag = type;
max.max = searchTree.getBalanceNumber();
}
if (Math.abs(searchTree.getBalanceNumber()) > Math.abs(max.max)) {
max.childTree = searchTree;
max.fatherTree = fatherTree;
max.flag = type;
max.max = searchTree.getBalanceNumber();
}
}
}
public static SearchTree buildTree(SearchTree searchTree, Integer value) {
if (Objects.isNull(searchTree)) {
searchTree = new SearchTree();
}
if (Objects.isNull(searchTree.getValue())) {
searchTree.setValue(value);
return searchTree;
}
if (value >= searchTree.getValue()) {
if (Objects.isNull(searchTree.getRightChild())) {
SearchTree searchTree1 = new SearchTree();
searchTree1.setValue(value);
searchTree.setRightChild(searchTree1);
} else {
buildTree(searchTree.getRightChild(), value);
}
} else {
if (Objects.isNull(searchTree.getLeftChild())) {
SearchTree searchTree1 = new SearchTree();
searchTree1.setValue(value);
searchTree.setLeftChild(searchTree1);
} else {
buildTree(searchTree.getLeftChild(), value);
}
}
return searchTree;
}
public static void main(String[] args) {
// int[] a = new int[]{2, 4, 1, 3, 5, 10, 9, 8};
int[] a = new int[]{2, 4, 1, 3, 5, 10, 9, 8, 6, 7};
SearchTree searchTree = new SearchTree();
for (int i = 0; i < a.length; i++) {
searchTree = buildTree(searchTree, a[i]);
countHeight(searchTree);
MaxNumber maxNumber = checkBalance(searchTree);
SearchTree searchTree1 = maxNumber.childTree;
if (Math.abs(searchTree1.getBalanceNumber()) >= 2) {
searchTree1 = rotate(searchTree1);
if (maxNumber.flag == 0) {
maxNumber.fatherTree = searchTree1;
searchTree = searchTree1;
} else if (maxNumber.flag == 1) {
maxNumber.fatherTree.setLeftChild(searchTree1);
} else if (maxNumber.flag == 2) {
maxNumber.fatherTree.setRightChild(searchTree1);
}
countHeight(searchTree);
}
}
System.out.println("最终为\n" + Json.toJson(searchTree));
}
}以序列2, 4, 1, 3, 5, 10, 9, 8, 6, 7为例，构造的平衡二叉搜索树结构为
{
"value": 4,
"left_child": {
"value": 2,
"left_child": {
"value": 1,
"left_child": null,
"right_child": null,
"balance_number": 0,
"height": 1
},
"right_child": {
"value": 3,
"left_child": null,
"right_child": null,
"balance_number": 0,
"height": 1
},
"balance_number": 0,
"height": 2
},
"right_child": {
"value": 8,
"left_child": {
"value": 6,
"left_child": {
"value": 5,
"left_child": null,
"right_child": null,
"balance_number": 0,
"height": 1
},
"right_child": {
"value": 7,
"left_child": null,
"right_child": null,
"balance_number": 0,
"height": 1
},
"balance_number": 0,
"height": 2
},
"right_child": {
"value": 10,
"left_child": {
"value": 9,
"left_child": null,
"right_child": null,
"balance_number": 0,
"height": 1
},
"right_child": null,
"balance_number": 1,
"height": 2
},
"balance_number": 0,
"height": 3
},
"balance_number": -1,
"height": 4
}到此，相信大家对“Java平衡二叉树怎么实现”有了更深的了解，不妨来实际操作一番吧！这里是亿速云网站，更多相关内容可以进入相关频道进行查询，关注我们，继续学习！
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