• java——二分搜索树 BST(递归、非递归)


    ~

    package Date_pacage;
    import java.util.Stack;
    import java.util.ArrayList;
    import java.util.LinkedList;
    import java.util.Queue;
    import java.util.Random;
    
    //二分搜索树
    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;
        private int size;
        public BST() {
            root = null;
            size = 0;
        }
        public int size() {
            return size;
        }
        public boolean isEmpty(){
            return size == 0;
        }
        //相等的元素不会重复添加
        public void add(E e) {
            root = add(root, e);
        }
        private Node add(Node node, E e) {
            if(node == null) {
                size ++;
                node = new Node(e);
            }
            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;
        }
        public boolean contains(E e) {
            return contains(root, e);
        }
        //搜索二分搜索树是否包含元素e
        private boolean contains(Node node, E e) {
            if(node == null)
                return false;
            if(e.compareTo(node.e) == 0) {
                return true;
            }else if(e.compareTo(node.e) > 0) {
                return contains(node.right, e);
            }else {
                return contains(node.left, e);
            }
        }
        //前序遍历
        public void preOrder() {
            preOrder(root);
        }
        private void preOrder(Node node) {
            if(node == null) {
                return;
            }
            System.out.println(node.e);
            preOrder(node.left);
            preOrder(node.right);
        }
        //非递归前序遍历:栈
        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);
                }
            }
        }
        //中序遍历
        public void inOrder() {
            preOrder(root);
        }
        private void inOrder(Node node) {
            if(node == null) {
                return;
            }
            inOrder(node.left);
            System.out.println(node.e);
            inOrder(node.right);
        }
        //后序遍历
        public void postOrder() {
            postOrder(root);
        }
        private void postOrder(Node node){
            if(node == null) {
                return;
            }
            postOrder(node.left);
            postOrder(node.right);
            System.out.println(node.e);
        }
        //非递归 层序遍历(广度优先遍历):队列
        public void levelOrder() {
            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);
                }
            }
        }
        //寻找二分搜索树的最小元素
        public E minimum() {
            if(size == 0) {
                throw new IllegalArgumentException("BST is empty!");
            }
            return minimum(root).e;
        }
        private Node minimum(Node node) {
            if(node.left == null) {
                return node;
            }
            return minimum(node.left);
        }
        public E removeMin() {
            E ret = minimum();
            root = removeMin(root);
            return ret;
        }
        //删除掉以node为跟的二分搜索树的最小a节点
        //返回删除节点后的新的二分搜索树的跟
        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;
        }
        @Override
        public String toString() {
            StringBuilder res = new StringBuilder();
            generateBSTString(root, 0, res);
            return res.toString();
        }
        private void generateBSTString(Node node, int depth, StringBuilder res) {
            if(node == null) {
                res.append(generateDepthString(depth) + "null
    ");
                return;
            }
            res.append(generateDepthString(depth) + node.e + "
    ");
            generateBSTString(node.left, depth+1, res);
            generateBSTString(node.right, depth+1, res);
        }
        
        private String generateDepthString(int depth) {
            StringBuilder res = new StringBuilder();
            for(int i = 0 ; i < depth ; i ++) {
                res.append("--");
            }
            return res.toString();
        }
        
        public static void main(String[] args) {
            BST<Integer> bst = new BST<>();
            Random random = new Random();
            int n = 1000;
            for(int i = 0 ; i < n ; i ++) {
                //从0 到10000的一个数
                bst.add(random.nextInt(10000));
            }
            ArrayList<Integer> nums = new ArrayList<>();
            while( !bst.isEmpty()) {
                nums.add(bst.removeMin());
            }
            System.out.println(nums);
            
        }
    }
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  • 原文地址:https://www.cnblogs.com/gaoquanquan/p/9846591.html
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