详解Java数据结构之平衡二叉树
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
- 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 }
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