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本篇內(nèi)容主要講解“Java平衡二叉樹怎么實(shí)現(xiàn)”，感興趣的朋友不妨來看看。本文介紹的方法操作簡(jiǎn)單快捷，實(shí)用性強(qiáng)。下面就讓小編來帶大家學(xué)習(xí)“Java平衡二叉樹怎么實(shí)現(xiàn)”吧!
簡(jiǎn)單來說，就是方便搜索的二叉樹，是一種具備特定結(jié)構(gòu)的二叉樹，即，對(duì)于節(jié)點(diǎn)n，其左子樹的所有節(jié)點(diǎn)的值都小于等于其值，其右子樹的所有節(jié)點(diǎn)的值都大于等于其值。
以序列2,4,1,3,5,10,9,8為例，如果以二叉搜索樹建樹的方式，我們建立出來的逐個(gè)步驟應(yīng)該為
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樹就是這么一個(gè)樣子。其實(shí)現(xiàn)代碼如下：
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));
}
}運(yùn)行的結(jié)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
}
}
}
}與我們的目標(biāo)結(jié)果是一致的。
好了，那我們本節(jié)就完畢了?？墒寝D(zhuǎn)過頭可能你也發(fā)現(xiàn)了，直接生成的這個(gè)二叉搜索樹似乎有點(diǎn)太長(zhǎng)了，層數(shù)有點(diǎn)太多了，一般來說，一個(gè)長(zhǎng)度為8的序列，四層結(jié)構(gòu)的二叉樹就可以表現(xiàn)出來了，這里卻使用了六層，顯然這樣的結(jié)果不盡人意，同時(shí)太深的層數(shù)，也增加了查找的時(shí)間復(fù)雜度。
這就給我們的樹提了要求，我們需要將目前構(gòu)造出來的樹平衡一下，讓這棵二叉搜索樹的左右子樹“重量”最好差不多。
首先需要掌握兩個(gè)概念
平衡因子
旋轉(zhuǎn)
平衡因子就是對(duì)于這棵二叉搜索樹的每個(gè)節(jié)點(diǎn)來說，其左子樹的高度減去右子樹的高度即為該節(jié)點(diǎn)的平衡因子，該數(shù)值能很快的辨別出該節(jié)點(diǎn)究竟是左子樹高還是右子樹高。在平衡二叉樹中規(guī)定，當(dāng)一個(gè)節(jié)點(diǎn)的平衡因子的絕對(duì)值大于等于2的時(shí)候，我們就認(rèn)為該節(jié)點(diǎn)不平衡，需要進(jìn)行調(diào)整。那么這種調(diào)整的手段稱之為節(jié)點(diǎn)與節(jié)點(diǎn)的旋轉(zhuǎn)，通俗來說，旋轉(zhuǎn)就是指的節(jié)點(diǎn)間的指向關(guān)系發(fā)生變化，在c語(yǔ)言中就是指針指向的切換。
在調(diào)用旋轉(zhuǎn)之前，我們需要判斷整棵樹是否平衡，即，這棵二叉搜索樹的所有平衡因子是否有絕對(duì)值大于等于2的，如果有，就找出最小的一棵子樹?？梢源_定的是，如果前一次二叉搜索樹是平衡的，那么此時(shí)如果加一個(gè)節(jié)點(diǎn)進(jìn)去，造成不平衡，那么節(jié)點(diǎn)從葉子開始回溯，找到的第一個(gè)大于等于2的節(jié)點(diǎn)勢(shì)必為最小不平衡子樹的根節(jié)點(diǎn)。
對(duì)于這棵最小不平衡的子樹，我們需要得到兩個(gè)值，即根節(jié)點(diǎn)的平衡因子a，以及左右子樹根節(jié)點(diǎn)中平衡因子絕對(duì)值較大者的平衡因子b。
我們可以將需要旋轉(zhuǎn)的類型抽象為以下四種：
1.左左型（正正型，即 a>0 && b>0）
左左型最后想要達(dá)到的目標(biāo)是第二個(gè)節(jié)點(diǎn)成為根節(jié)點(diǎn)，第一個(gè)節(jié)點(diǎn)成為第二個(gè)節(jié)點(diǎn)的右節(jié)點(diǎn)。
所以用偽代碼展示就是（設(shè)a1，a2，a3分別為圖里面從上到下的三個(gè)節(jié)點(diǎn)）
a2的右子樹 = (合并(a2的右子樹,a1的右子樹) + a1頂點(diǎn)值) 一起構(gòu)成的二叉搜索樹;
返回 a2
2.左右型（正負(fù)型，即 a>0 && b<0）
設(shè)a1，a2，a3分別為圖里面從上到下的三個(gè)節(jié)點(diǎn)
首先應(yīng)該通過將a3和a2調(diào)換上下位置，使之變成左左型，然后再調(diào)用左左型的方法就完成了。
從左右型調(diào)換成左左型，即將a2及其左子樹成為a3左子樹的一部分，然后將a1的左子樹置為a3即可。
偽代碼如下：
a3的左子樹 = a2及其左子樹與a3的左子樹合并成的一棵二叉搜索樹;
a1的左子樹 = a3;
3.右右型（負(fù)負(fù)型，即 a<0 && b<0）
設(shè)a1，a2，a3分別為圖里面從上到下的三個(gè)節(jié)點(diǎn)
右右型與左左型類似，要達(dá)到的目的就是a1成為a2左子樹的一部分，偽代碼為：
a2的左子樹 = （合并a2的左子樹和a1的左子樹）+ a1頂點(diǎn)的值構(gòu)成的二叉搜索樹;
返回a2
4.右左型（負(fù)正型，即 a<0 && b>0）
設(shè)a1，a2，a3分別為圖里面從上到下的三個(gè)節(jié)點(diǎn)
右左型需要先轉(zhuǎn)換成右右型，然后在調(diào)用右右型的方法即可。
從右左型到右右型，即需要將a2及其右子樹成為a3右子樹的一部分，然后將a1的右子樹置為a3即可。
偽代碼如下：
a3的右子樹 = a2及其右子樹與a3的右子樹合并成的一棵二叉搜索樹;
a1的右子樹 = a3;
從上面的分析可以得出，我們不僅僅需要實(shí)現(xiàn)旋轉(zhuǎn)的方法，還需要實(shí)現(xiàn)合并二叉樹等方法，這些方法都是基礎(chǔ)方法，讀者需要確保會(huì)快速寫出來。
請(qǐng)讀者朋友們根據(jù)上面的內(nèi)容，先嘗試寫出集中平衡化的方法。
平衡二叉搜索樹建樹，需要在二叉搜索樹建樹的基礎(chǔ)上加上平衡的過程，即子樹之間指針轉(zhuǎn)換的問題，同時(shí)，由于這種指針轉(zhuǎn)換引起的子樹的子樹也會(huì)產(chǎn)生不平衡，所以上面提到的四種旋轉(zhuǎn)調(diào)整方式都是遞歸的。
首先，先構(gòu)建節(jié)點(diǎn)基礎(chǔ)結(jié)構(gòu)：
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，左子樹，右子樹
這是計(jì)算二叉搜索樹中每個(gè)平衡因子的基礎(chǔ)，我們?cè)O(shè)最低層為高度1，則計(jì)算節(jié)點(diǎn)高度的代碼為：
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();
}這里有個(gè)半動(dòng)態(tài)規(guī)劃的結(jié)論：當(dāng)前節(jié)點(diǎn)的高度，等于左右子樹的最大高度+1；這里的寫法有點(diǎn)樹形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);
}
}本質(zhì)上講，平衡因子就是左子樹高度減去右子樹高度，注意這里左右子樹都有可能不存在，所以加入了一堆特判。
判斷當(dāng)前二叉樹是否平衡
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是左子樹，2代表childTree是右子樹
}
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類是為了保存第一棵不平衡的子樹而存在的結(jié)構(gòu)，為了使這棵子樹平衡之后能重新回到整棵樹中，需要在MaxNumber中存儲(chǔ)當(dāng)前子樹父節(jié)點(diǎn)，同時(shí)標(biāo)明當(dāng)前子樹是父節(jié)點(diǎn)的左子樹還是右子樹，還是本身。
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);
}
}
/**
* 合并兩棵二叉搜索樹
*
* @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;
}將一棵樹轉(zhuǎn)成數(shù)字集合，然后通過建樹的方式建到另外一棵樹上即可。
1.左左型旋轉(zhuǎn)
/**
* 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.右右型旋轉(zhuǎn)
/**
* 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.左右型旋轉(zhuǎn)
/**
* 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.右左型旋轉(zhuǎn)
/**
* 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;
}旋轉(zhuǎn)調(diào)用函數(shù)：
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是左子樹，2代表childTree是右子樹
}
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);
}
}
/**
* 合并兩棵二叉搜索樹
*
* @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問題
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("最終為\n" + Json.toJson(searchTree));
}
}以序列2, 4, 1, 3, 5, 10, 9, 8, 6, 7為例，構(gòu)造的平衡二叉搜索樹結(jié)構(gòu)為
{
"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，相信大家對(duì)“Java平衡二叉樹怎么實(shí)現(xiàn)”有了更深的了解，不妨來實(shí)際操作一番吧！這里是億速云網(wǎng)站，更多相關(guān)內(nèi)容可以進(jìn)入相關(guān)頻道進(jìn)行查詢，關(guān)注我們，繼續(xù)學(xué)習(xí)！
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