0 课程地址
https://coding.imooc.com/lesson/207.html#mid=14352
1 重点关注
1.1 本节要点
前两节新增节点后维护平衡的方法直接拿过来用即可
1.2 修改删除元素维护平衡后需要注意的情况
a removMin方法有可能打破平衡
b node 查找到后为 互斥关系
c 假如删除后节点为空,会有空指针异常
2 课程内容
2.1 AVL删除元素代码实现
见3.1
2.2 AVL删除元素代码实现测试
见3.1
3 Coding
3.1 coding
案例:傲慢与偏见 统计单词
- 关键代码
private Node remove(Node node, K key){ //终止条件1 基本判断不到,因为已经判断了containskey if(node==null){ return null; } Node retNode; //递归 if(key.compareTo(node.key)<0){ node.left = remove(node.left,key); retNode = node; }else if(key.compareTo(node.key)>0){ node.right = remove(node.right,key); retNode = node; }else{ //已找到要删除的元素 //1 如果只有左子节点或只有右子节点,则直接将子节点替换 if(node.left==null){ retNode = node.right; size--; }else if(node.right==null){ retNode = node.left; size--; }else{ //2 如果有左子节点和右子节点,则寻找前驱或后继 对当前节点替换掉 Node nodeMain = findMin(node.right); nodeMain.right = remove(node.right,nodeMain.key);//这块要么用removMin方法(添加维护平衡的代码),要么复用remove方法(删除以node右子树为根的二叉树的最小值) nodeMain.left = node.left; node.left = node.right = null; retNode = nodeMain; size--; } } if(retNode==null){ return null; } retNode.height = 1+Math.max(getHeight(retNode.left),getHeight(retNode.right)); int balanceFactor = getBalanceFactor(retNode); //左左情况 if(balanceFactor>1&&getBalanceFactor(retNode.left)>=0){ return rightRotate(retNode); //右右情况 }else if(balanceFactor<-1&&getBalanceFactor(retNode.right)<=0){ return leftRotate(retNode); //左右情况 }else if(balanceFactor>1&&getBalanceFactor(retNode.left)<0){ //先对左子节点左旋转,然后整体右旋转 retNode.left = leftRotate(retNode.left); return rightRotate(retNode); //右左情况 }else if(balanceFactor<-1&&getBalanceFactor(retNode.right)>0){ retNode.right = rightRotate(retNode.right); return leftRotate(retNode); } return retNode; }
- 主类AVLTree:
package com.company; import java.util.ArrayList; import java.util.List; public class AVLTree<K extends Comparable<K>,V> { //1 定义Node class Node{ private K key; private V value; private Node left,right; private int height; public Node(K key, V value){ this.key = key; this.value = value; this.left = null; this.right = null; this.height = 1; } @Override public String toString() { final StringBuffer sb = new StringBuffer("Node{"); sb.append("key=").append(key); sb.append(", value=").append(value); sb.append('}'); return sb.toString(); } } //2 定义属性 private int size; private Node root; /** * getHeight * @author weidoudou * @date 2023/4/1 11:34 * @param node 请添加参数描述 * @return int **/ private int getHeight(Node node){ if(null==node){ return 0; } return node.height; } /** * getBalanceFactor * @author weidoudou * @date 2023/4/1 11:36 * @param node 请添加参数描述 * @return int **/ private int getBalanceFactor(Node node){ if(null==node){ return 0; } return getHeight(node.left) - getHeight(node.right); } /** * 无参构造函数 * @author weidoudou * @date 2023/1/1 11:09 * @return null **/ public AVLTree(){ this.size = 0; this.root = null; } public boolean isEmpty() { return size==0?true:false; } public int getSize() { return size; } //3 定义包含函数 private Node containsKey(K key,Node node){ //结束条件 if(null==node){ return null; } //循环条件 if(key.compareTo(node.key)<0){ return containsKey(key,node.left); }else if(key.compareTo(node.key)>0){ return containsKey(key, node.right); }else{//key.compareTo(node.key)=0 其实这个也是结束条件 return node; } } public boolean contains(K key) { return containsKey(key,root)==null?false:true; } public V get(K key) { Node node = containsKey(key,root); if(null!=node){ return node.value; } return null; } //3 递归,添加元素 public Node add(K key,V value,Node node){ //3.1 终止条件 //3.1.1 要插入的元素和二叉树原有节点相同,这个不用判断,因为已经调了containsKey方法判断了 if(node==null){ size++; return new Node(key,value); } //3.1.2 最终插入左孩子 if(key.compareTo(node.key)<0 ){ node.left = add(key,value,node.left); }else if(key.compareTo(node.key)>0){ node.right = add(key,value,node.right); }else{ node.value = value; } node.height = 1+Math.max(getHeight(node.left),getHeight(node.right)); int balanceFactor = getBalanceFactor(node); /*if(Math.abs(balanceFactor)>1){ System.out.println("unbalanced:"+balanceFactor); }*/ //左左情况 if(balanceFactor>1&&getBalanceFactor(node.left)>=0){ return rightRotate(node); //右右情况 }else if(balanceFactor<-1&&getBalanceFactor(node.right)<=0){ return leftRotate(node); //左右情况 }else if(balanceFactor>1&&getBalanceFactor(node.left)<0){ //先对左子节点左旋转,然后整体右旋转 node.left = leftRotate(node.left); return rightRotate(node); //右左情况 }else if(balanceFactor<-1&&getBalanceFactor(node.right)>0){ node.right = rightRotate(node.right); return leftRotate(node); } return node; } // 对节点y进行向右旋转操作,返回旋转后新的根节点x // y x // / \ / \ // x T4 向右旋转 (y) z y // / \ - - - - - - - -> / \ / \ // z T3 T1 T2 T3 T4 // / \ // T1 T2 /** * 右旋转 * 1 旋转 * 2 变更高度 * @author weidoudou * @date 2023/4/14 7:27 * @param y 请添加参数描述 * @return com.company.AVLTree<K,V>.Node **/ private Node rightRotate(Node y){ //1 右旋转 Node x = y.left; Node T3 = x.right; x.right = y; y.left = T3; //2 变更高度 y.height = Math.max(getHeight(y.left),getHeight(y.right))+1; x.height = Math.max(getHeight(x.left),getHeight(x.left))+1; return x; } // 对节点y进行向左旋转操作,返回旋转后新的根节点x // y x // / \ / \ // T1 x 向左旋转 (y) y z // / \ - - - - - - - -> / \ / \ // T2 z T1 T2 T3 T4 // / \ // T3 T4 private Node leftRotate(Node y) { Node x = y.right; Node T2 = x.left; // 向左旋转过程 x.left = y; y.right = T2; // 更新height y.height = Math.max(getHeight(y.left), getHeight(y.right)) + 1; x.height = Math.max(getHeight(x.left), getHeight(x.right)) + 1; return x; } public void add(K key, V value) { root = add(key,value,root); /*Node node = containsKey(key,root); //未找到,插值 if(node==null){ //2.1 考虑特殊情况,如果是第一次调用,root为null if(root==null){ root = new Node(key,value); size++; }else{ //2.2 添加递归方法 add(key,value,root); } }else{ node.value = value; }*/ //找到后,更新值 } public void set(K key, V value) { Node node = containsKey(key,root); if(node == null){ throw new IllegalArgumentException("要修改的值不存在"); } node.value = value; } private Node remove(Node node, K key){ //终止条件1 基本判断不到,因为已经判断了containskey if(node==null){ return null; } Node retNode; //递归 if(key.compareTo(node.key)<0){ node.left = remove(node.left,key); retNode = node; }else if(key.compareTo(node.key)>0){ node.right = remove(node.right,key); retNode = node; }else{ //已找到要删除的元素 //1 如果只有左子节点或只有右子节点,则直接将子节点替换 if(node.left==null){ retNode = node.right; size--; }else if(node.right==null){ retNode = node.left; size--; }else{ //2 如果有左子节点和右子节点,则寻找前驱或后继 对当前节点替换掉 Node nodeMain = findMin(node.right); nodeMain.right = remove(node.right,nodeMain.key);//这块要么用removMin方法(添加维护平衡的代码),要么复用remove方法(删除以node右子树为根的二叉树的最小值) nodeMain.left = node.left; node.left = node.right = null; retNode = nodeMain; size--; } } if(retNode==null){ return null; } retNode.height = 1+Math.max(getHeight(retNode.left),getHeight(retNode.right)); int balanceFactor = getBalanceFactor(retNode); //左左情况 if(balanceFactor>1&&getBalanceFactor(retNode.left)>=0){ return rightRotate(retNode); //右右情况 }else if(balanceFactor<-1&&getBalanceFactor(retNode.right)<=0){ return leftRotate(retNode); //左右情况 }else if(balanceFactor>1&&getBalanceFactor(retNode.left)<0){ //先对左子节点左旋转,然后整体右旋转 retNode.left = leftRotate(retNode.left); return rightRotate(retNode); //右左情况 }else if(balanceFactor<-1&&getBalanceFactor(retNode.right)>0){ retNode.right = rightRotate(retNode.right); return leftRotate(retNode); } return retNode; } private Node findMin(Node node){ //1 终止条件 if(node.left==null){ return node; } //2 递归 return findMin(node.left); } private Node removMin(Node node){ //终止条件 if(node.left==null){ Node rightNode = node.right; node.right = null; return rightNode; } //递归 node.left = removMin(node.left); return node; } /** * 删除任意元素 若删除元素节点下只有一个节点直接接上即可,若有两个节点,则找前驱或后继,本节找前驱 * @author weidoudou * @date 2023/1/1 11:52 * @param key 请添加参数描述 * @return V **/ public V remove(K key) { Node node = containsKey(key,root); if(node != null){ root = remove(root, key); return node.value; } return null; } //1 校验二分搜索树(中序遍历参考之前的中序遍历一节) public boolean judgeBST(){ List<K> list = new ArrayList<>(); inOrder(root,list); for(int i=1;i<list.size();i++){ if(list.get(i-1).compareTo(list.get(i))>0){ return false; } } return true; } private void inOrder(Node node, List<K> list){ if(node==null){ return; } inOrder(node.left,list); list.add(node.key); inOrder(node.right,list); } //2 校验平衡二叉树 public boolean judgeBalance(){ return judgeBalance(root); } private boolean judgeBalance(Node node){ if(node == null){ return true; } int balanceFactor = getBalanceFactor(node); if(Math.abs(balanceFactor)>1){ return false; } return judgeBalance(node.left)&&judgeBalance(node.right); } public static void main(String[] args) { System.out.println("Pride and Prejudice"); ArrayList<String> words = new ArrayList<>(); if(FileOperation.readFile("pride-and-prejudice.txt",words)){ System.out.println("Total words: "+words.size()); AVLTree<String,Integer> avlTree = new AVLTree<>(); for(String word:words){ if(avlTree.contains(word)){ avlTree.set(word,avlTree.get(word)+1); }else{ avlTree.add(word,1); } } System.out.println("Total different words:"+avlTree.getSize()); System.out.println("judge BST:"+avlTree.judgeBST()); System.out.println("judge Balance:"+avlTree.judgeBalance()); for(String word:words){ avlTree.remove(word); if(!avlTree.judgeBalance()||!avlTree.judgeBST()){ System.out.println("平衡二叉树删除元素后不满足二叉树或者平衡二叉树性质"); } } } } }
- 测试结果:
Pride and Prejudice Total words: 125901 Total different words:6530 judge BST:true judge Balance:true Process finished with exit code 0
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