数据结构(底层实现+Java版本)
//二分搜索树(数据具有可比性)
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
public class BST<E extends Comparable<E>> {
private class Node {
public E e;
public Node left, right;
public Node(E e) {
this.e = e;
left = null;
right = null;
}
}
private Node root; //根节点
private int size; //树中的元素个数
public BST() {
root = null;
size = 0;
}
public int size() {
return size;
}
public boolean isEmpty() {
return size == 0;
}
//添加(递归算法)
public void add(E e) {
root = add(root, e);
}
//递归的方法(1.递归终止条件的判断,2.递归调用)
private Node add(Node node, E e) {
//递归终止条件的判断
if (node == null) {
return new Node(e); //如果根节点是空的话,直接返回一个新的空节点
}
//递归调用
if (e.compareTo(node.e) < 0) {
node.left = add(node.left, e);
} else if (e.compareTo(node.e) > 0) {
node.right = add(node.right, e);
}
return node;
}
//查询
public boolean contains(E e) {
return contains(root, e);
}
private boolean contains(Node node, E e) {
if (node == null) { //二分搜索树直接为空
return false;
}
//递归终止
if (e.compareTo(node.e) == 0) { //做减法
return true;
} else if (e.compareTo(node.e) < 0) {
return contains(node.left, e);
} else {
return contains(node.right, e);
}
}
//前序遍历
public void preOrder() {
preOrder(root);
}
private void preOrder(Node node) {
if (node == null) {
return;
}
System.out.println(node.e); //输出表示已经访问了
preOrder(node.left);
preOrder(node.right);
}
//中序遍历(从小到大的顺序)
public void inOrder() {
inOrder(root);
}
private void inOrder(Node node) {
if (node == null) {
return;
}
preOrder(node.left);
System.out.println(node.e); //输出表示已经访问了
preOrder(node.right);
}
//后序遍历
public void postOrder() {
postOrder(root);
}
private void postOrder(Node node) {
if (node == null) {
return;
}
preOrder(node.left);
preOrder(node.right);
System.out.println(node.e); //输出表示已经访问了
}
//前序遍历的非递归写法
public void preOrderNR() {
Stack<Node> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
Node cur = stack.pop();
System.out.println(cur.e);
if (cur.right != null) {
stack.push(cur.right);
}
if (cur.left != null) {
stack.push(cur.left);
}
}
}
//层序遍历(广度优先遍历)
public void levalOrder() {
Queue<Node> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
Node cur = q.remove();
System.out.println(cur.e);
if (cur.left != null) {
q.add(cur.left);
}
if (cur.right != null) {
q.add(cur.right);
}
}
}
//获得最小值(二分搜索树)
public E minmum() {
if (size == 0) {
throw new IllegalArgumentException("二分搜索树为空");
}
return minmum(root).e;
}
private Node minmum(Node node) {
if (node.left == null) { //递归终止条件的判断
return node;
}
//递归调用
return minmum(node.left);
}
//获得二叉树的最大值
public E maxmum() {
if (size == 0) {
throw new IllegalArgumentException("二分搜索树为空");
}
return maxmum(root).e;
}
private Node maxmum(Node node) {
if (node.right == null) { //递归终止条件的判断
return node;
}
//递归调用
return minmum(node.right);
}
//删除二分搜索树的最小值(返回最小值)
public E removeMin() {
E ret = minmum();
root = removeMin(root);
return ret;
}
private Node removeMin(Node node) {
if (node.left == null) {
Node rightNode = node.right;
node.right = null;
size--;
return rightNode;
}
node.left = removeMin(node.left);
return node;
}
//删除二分搜索树的最大值(并且返回根节点,就是删除得那个节点)
public E removeMax() {
E ret = maxmum();
root = removeMax(root);
return ret;
}
//递归写法,并且返回删除的节点
private Node removeMax(Node node) {
if (node.right == null) {
Node leftNode = node.left;
node.left = null; //让node与二分搜索树想离
size--;
return leftNode;
}
node.right = removeMin(node.right);
return node;
}
//在二分搜索树中删除任意节点(重点)
public void remove(E e) {
root = remove(root, e);
}
private Node remove(Node node, E e) {
if (node == null) {
return null;
}
if (e.compareTo(node.e) < 0) {
node.left = remove(node.left, e);
return node;
} else if (e.compareTo(node.e) > 0) {
node.right = remove(node.right, e);
return node;
} else {
if (node.left == null) { //待删除节点的左子树为空
Node rightNode = node.right; //用于返回的新的根节点
node.right = null; //让该节点与整个二分搜索树脱节(这是一种写的方式)
size--;
return rightNode;
}
if (node.right == null) { //待删除节点的右子树为空
Node leftNode = node.left; //用于返回的新的根节点
node.left = null; //让该节点与整个二分搜索树脱节(这是一种写的方式)
size--;
return leftNode;
}
//删除的节点左右都不为空(待删除节点的右子树的最小节点作为新的根节点,或者左子树的最大节点作为新的根节点)。
Node n = minmum(node.right);
n.right = removeMin(node.right);
n.left = node.left;
node.left = node.right = null;
return n;
}
}
}