一.概念

二叉搜索树又称二叉排序树，具有以下性质：

• 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

注意：二叉搜索树中序遍历的结果是有序的

二、基本操作

1.查找元素

思路：二叉搜索树的左子树永远是比根节点小的，而它的右子树则都是比根节点大的值。当前节点比要找的大就往左走，当前元素比要找的小就往右走

	public Node search(int key) {if(root == null) {return null;}Node cur = root;while (cur != null) {if(cur.val == key) {return cur;}else if(cur.val > key) {cur = cur.left;}else{cur = cur.right;}}return null;}

2.插入元素

如果是空树直接把元素插入root位置就好了
思路：因为是二叉搜索树就不能插入重复的元素了，且每次插入都是插入到叶子节点的位置。定义一个 cur 从root开始，插入的元素比当前位置元素小就往左走，比当前位置元素大就往右走，直到为空，所以就需要再定义一个变量parent 记住 cur 的前面的位置。
最后再判断插入到parent 的左子树还是右子树位置

代码实现：

	public boolean insert(int key) {Node node = new Node(key);if(root == null) {this.root = node;return true;}Node parent = null;Node cur = root;while (cur != null) {if(cur.val == key) {//有相同的元素直接returnreturn false;}else if(cur.val > key) {parent = cur;cur = cur.left;}else{parent = cur;cur = cur.right;}}if (parent.val > key) {parent.left = node;}else{parent.right = node;}return true;}

3.删除元素

删除元素是一个比较难的点，要考虑到很多种情况

1. cur.left == null

   1. cur 是 root，则 root = cur.right
   2. cur 不是 root，cur 是 parent.left，则 parent.left = cur.right
   3. cur 不是 root，cur 是 parent.right，则 parent.right = cur.right

2. cur.right == null

   1. cur 是 root，则 root = cur.left
   2. cur 不是 root，cur 是 parent.left，则 parent.left = cur.left
   3. cur 不是 root，cur 是 parent.right，则 parent.right = cur.left

3. cur.left != null && cur.right != null
   采用替罪羊的方式删除

   1. 找到要删除节点，右树最左边的节点或者找到左树最右边的节点，替换这个两个节点的val值。
   2. 这样才能保证，删除后左树一定比根节点小，右树一定比根节点大
      

	public boolean remove(int key) {if(this.root == null) {return false;}Node parent = null;Node cur = this.root;while (cur != null) {if(cur.val == key) {removeKey(parent,cur);return true;}else if(cur.val < key) {parent = cur;cur = cur.right;}else{parent = cur;cur = cur.left;}}return false;}public void removeKey(Node parent,Node cur) {if(cur.left == null) {if(cur == this.root) {this.root = this.root.right;}else if(cur == parent.left) {parent.left = cur.right;}else{parent.right = cur.right;}}else if(cur.right == null) {if(this.root == cur) {this.root = this.root.left;}else if(cur == parent.left) {parent.left = cur.left;}else{parent.right = cur.left;}}else{//左右都不为空的情况Node targetParent = cur;Node target = cur.right;while (target.left != null) {targetParent = target;target = target.left;}cur.val = target.val;if(targetParent.left == target) {targetParent.left = target.right;}else{targetParent.right = target.right;}}}

4.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：

最好情况：二叉搜索树为完全二叉树，其平均比较次数为 O(log${}_2$n)

最坏情况：二叉搜索树退化为单支树，其平均比较次数为：O(n)

所有代码：

public class BinarySearchTree {public static class Node {int val;Node left;Node right;public Node(int val) {this.val = val;}}public Node root = null;/*** 查找某个节点* @param key*/public Node search(int key) {if(root == null) {return null;}Node cur = root;while (cur != null) {if(cur.val == key) {return cur;}else if(cur.val > key) {cur = cur.left;}else{cur = cur.right;}}return null;}/*** 插入元素* @param key* @return*/public boolean insert(int key) {Node node = new Node(key);if(root == null) {this.root = node;return true;}Node parent = null;Node cur = root;while (cur != null) {if(cur.val == key) {//有相同的元素直接returnreturn false;}else if(cur.val > key) {parent = cur;cur = cur.left;}else{parent = cur;cur = cur.right;}}if (parent.val > key) {parent.left = node;}else{parent.right = node;}return true;}/*** 删除元素* @param key*/public boolean remove(int key) {if(this.root == null) {return false;}Node parent = null;Node cur = this.root;while (cur != null) {if(cur.val == key) {removeKey(parent,cur);return true;}else if(cur.val < key) {parent = cur;cur = cur.right;}else{parent = cur;cur = cur.left;}}return false;}public void removeKey(Node parent,Node cur) {if(cur.left == null) {if(cur == this.root) {this.root = this.root.right;}else if(cur == parent.left) {parent.left = cur.right;}else{parent.right = cur.right;}}else if(cur.right == null) {if(this.root == cur) {this.root = this.root.left;}else if(cur == parent.left) {parent.left = cur.left;}else{parent.right = cur.left;}}else{Node targetParent = cur;Node target = cur.right;while (target.left != null) {targetParent = target;target = target.left;}cur.val = target.val;if(targetParent.left == target) {targetParent.left = target.right;}else{targetParent.right = target.right;}}}
}