java

关注公众号 jb51net

关闭
首页 > 软件编程 > java > Java平衡二叉树

详解Java数据结构之平衡二叉树

作者：炒鸡辣鸡123

平衡二叉树（Balanced Binary Tree）又被称为AVL树（有别于AVL算法），且具有以下性质：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。本文将详解介绍一下平衡二叉树的原理与实现，需要的可以参考一下

什么是二叉搜索树

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。

平衡二叉搜索树

首先需要掌握两个概念

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程序

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

计算每个节点的高度

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。

计算每个节点的平衡因子

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。

合并二叉树

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。

旋转调整函数

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;
}

整体代码

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
}

以上就是详解Java数据结构之平衡二叉树的详细内容，更多关于Java平衡二叉树的资料请关注脚本之家其它相关文章！

您可能感兴趣的文章:
阅读全文