JavaSE 搜索树

1 概念

二叉搜索树又称二叉排序树，它是一棵空树，或者是具有以下性质的二叉树:

若它的左子树不为空，则左子树上所有节点的值都小于根节点的值；
若它的右子树不为空，则右子树上所有节点的值都大于根节点的值。
它的左右子树也分别为二叉搜索树

你会发现它中序遍历的结果就是有序的。
如下图所示就是一颗二叉搜索树：
在这里插入图片描述

2 操作

2.1 查找

在这里插入图片描述
具体实现代码示例如下所示：

package bstree;class BinarySearchTree{static class BSNode{public int val;public BSNode left;public BSNode right;public BSNode(int val) {this.val = val;}}public BSNode root = null;public BSNode search(int val){if(root == null) return null;BSNode cur = root;while (cur != null){if(cur.val == val){return cur;}else if(cur.val > val){cur = cur.left;}else{cur = cur.right;}}return null;}}public class TestDemo {public static void main(String[] args) {}
}

2.2 插入

如果树为空树，即根 == null，直接插入。
如果树不是空树，按照查找逻辑确定插入位置，插入新结点。

具体实现代码示例如下所示：

package bstree;class BinarySearchTree{static class BSNode{public int val;public BSNode left;public BSNode right;public BSNode(int val) {this.val = val;}}public BSNode root = null;public BSNode search(int val){if(root == null) return null;BSNode cur = root;while (cur != null){if(cur.val == val){return cur;}else if(cur.val > val){cur = cur.left;}else{cur = cur.right;}}return null;}public boolean insert(int val){BSNode bsNode = new BSNode(val);if(root == null){root = bsNode;return true;}BSNode cur = root;BSNode parent = null;while(cur != null){if(cur.val == val){return false;}else if(cur.val > val){parent = cur;cur = cur.left;}else{parent = cur;cur = cur.right;}}if(parent.val < val){parent.right = bsNode;}else{parent.left = bsNode;}return true;}}public class TestDemo {public static void preOrder(BinarySearchTree.BSNode root){if(root == null){return;}System.out.print(root.val+" ");preOrder(root.left);preOrder(root.right);}public static void inOrder(BinarySearchTree.BSNode root){if(root == null){return;}inOrder(root.left);System.out.print(root.val+" ");inOrder(root.right);}public static void main(String[] args) {BinarySearchTree binarySearchTree = new BinarySearchTree();binarySearchTree.insert(4);binarySearchTree.insert(3);binarySearchTree.insert(1);binarySearchTree.insert(15);binarySearchTree.insert(11);preOrder(binarySearchTree.root);System.out.println();inOrder(binarySearchTree.root);System.out.println();try {BinarySearchTree.BSNode ret = binarySearchTree.search(4);System.out.println(ret.val);}catch (NullPointerException e){System.out.println("没有找到当前的节点..........");e.printStackTrace();}}
}

2.3 删除

前提是删除这个节点之后，整棵树还是一棵二叉搜索树。
删除思路：
设待删除结点为 cur, 待删除结点的双亲结点为 parent。
则分为下面三种情况：

cur.left == null

cur 是 root，则 root = cur.right
cur 不是 root，cur 是 parent.left，则 parent.left = cur.right
cur 不是 root，cur 是 parent.right，则 parent.right = cur.right

cur.right == null

cur 是 root，则 root = cur.left
cur 不是 root，cur 是 parent.left，则 parent.left = cur.left
cur 不是 root，cur 是 parent.right，则 parent.right = cur.left

cur.left != null && cur.right != null

如果像原来那样删除，那么一个节点就会出现两个父亲节点？
需要使用替换法进行删除，即在它的右子树中寻找中序下的第一个结点(关键值最小)，用它的值填补到被删除节点中，再来处理该结点的删除问题。
我们怎么知道放谁上去？
当前需要删除的节点的左边找最大的，右边找最小的。

具体实现代码示例如下所示：

package bstree;class BinarySearchTree {static class BSNode {public int val;public BSNode left;public BSNode right;public BSNode(int val) {this.val = val;}}public BSNode root = null;public BSNode search(int val) {if (root == null) return null;BSNode cur = root;while (cur != null) {if (cur.val == val) {return cur;} else if (cur.val > val) {cur = cur.left;} else {cur = cur.right;}}return null;}public boolean insert(int val) {BSNode bsNode = new BSNode(val);if (root == null) {root = bsNode;return true;}BSNode cur = root;BSNode parent = null;while (cur != null) {if (cur.val == val) {return false;} else if (cur.val > val) {parent = cur;cur = cur.left;} else {parent = cur;cur = cur.right;}}if (parent.val < val) {parent.right = bsNode;} else {parent.left = bsNode;}return true;}public void remove(int val) {if (root == null) return;BSNode cur = root;BSNode parent = null;while(cur !=null){if (cur.val == val) {removeNode(parent,cur,val);} else if (cur.val < val) {parent = cur;cur = cur.right;} else {parent = cur;cur = cur.left;}}}public void removeNode(BSNode parent,BSNode cur,int val){if(cur.left == null){if(cur == root){root = cur.right;}else if(parent.left == cur){parent.left = cur.right;}else if(parent.right == cur){parent.right = cur.right;}}else if(cur.right == null){if(cur == root){root = cur.left;}else if(parent.left == cur){parent.left = cur.left;}else if(parent.right == cur){parent.right = cur.left;}}else{//这里采取的是右边找最小的方法BSNode targetParent = cur;BSNode target = cur.right;while(target.left != null){targetParent = target;target = target.left;}//target指向的节点就是 右边的最小值cur.val = target.val;if(target == targetParent.left){targetParent.left = target.right;}else{targetParent.right = target.right;}}}}public class TestDemo {public static void preOrder(BinarySearchTree.BSNode root){if(root == null){return;}System.out.print(root.val+" ");preOrder(root.left);preOrder(root.right);}public static void inOrder(BinarySearchTree.BSNode root){if(root == null){return;}inOrder(root.left);System.out.print(root.val+" ");inOrder(root.right);}public static void main(String[] args) {BinarySearchTree binarySearchTree = new BinarySearchTree();binarySearchTree.insert(4);binarySearchTree.insert(3);binarySearchTree.insert(1);binarySearchTree.insert(15);binarySearchTree.insert(11);preOrder(binarySearchTree.root);System.out.println();inOrder(binarySearchTree.root);System.out.println();binarySearchTree.remove(15);System.out.println("=============删除===============");preOrder(binarySearchTree.root);System.out.println();inOrder(binarySearchTree.root);System.out.println();try {BinarySearchTree.BSNode ret = binarySearchTree.search(4);System.out.println(ret.val);}catch (NullPointerException e){System.out.println("没有找到当前的节点..........");e.printStackTrace();}}
}

3 性能分析

插入和删除操作都必须先查找，查找效率代表了二叉搜索树中各个操作的性能。
对有n个结点的二叉搜索树，若每个元素查找的概率相等，则二叉搜索树平均查找长度是结点在二叉搜索树的深度的函数，即结点越深，则比较次数越多。
但对于同一个关键码集合，如果各关键码插入的次序不同，可能得到不同结构的二叉搜索树：
在这里插入图片描述
最优情况下： 二叉搜索树为完全二叉树，其比较次数为：O(log2^n)
最差情况下: 二叉搜索树退化为单支树，其比较次数为：O(n)。
问题：如果退化成单支树，二叉搜索树的性能就失去了。那能否进行改进，不论按照什么次序插入关键码，都可以是二叉搜索树的性能最佳？