java二叉查找树的实现代码

package 查找;import edu.princeton.cs.algs4.Queue;import edu.princeton.cs.algs4.StdOut;public class BST<Key extends Comparable<Key>, Value> {  private class Node {    private Key key; // 键    private Value value;// 值    private Node left, right; // 指向子树的链接    private int n; // 以该节点为根的子树中的节点总数    public Node(Key key, Value val, int n) {      this.key = key;      this.value = val;      this.n = n;    }  }  private Node root;  public int size() {    return size(root);  }  private int size(Node x) {    if (x == null)      return 0;    else      return x.n;  }  /**   * 如果树是空的，则查找未命中 如果被查找的键小于根节点，则在左子树中继续查找 如果被查找的键大于根节点，则在右子树中继续查找   * 如果被查找的键和根节点的键相等，查找命中   *    * @param key   * @return   */  public Value get(Key key) {    return get(root, key);  }  private Value get(Node x, Key key) {    if (x == null)      return null;    int cmp = key.compareTo(x.key);    if (cmp < 0)      return get(x.left, key);    else if (cmp > 0)      return get(x.right, key);    else      return x.value;  }  /**   * 二叉查找树的一个很重要的特性就是插入的实现难度和查找差不多。 当查找到一个不存在与树中的节点（null）时，new 新节点，并将上一路径指向该节点   *    * @param key   * @param val   */  public void put(Key key, Value val) {    root = put(root, key, val);  }  private Node put(Node x, Key key, Value val) {    if (x == null)      return new Node(key, val, 1);    int cmp = key.compareTo(x.key);    if (cmp < 0)      x.left = put(x.left, key, val);    else if (cmp > 0)      x.right = put(x.right, key, val);    else      x.value = val;    x.n = size(x.left) + size(x.right); // 要及时更新节点的子树数量    return x;  }  public Key min() {    return min(root).key;  }  private Node min(Node x) {    if (x.left == null)      return x;    return min(x.left);  }  public Key max() {    return max(root).key;  }  private Node max(Node x) {    if (x.right == null)      return x;    return min(x.right);  }  /**   * 向下取整：找出小于等于该键的最大键   *    * @param key   * @return   */  public Key floor(Key key) {    Node x = floor(root, key);    if (x == null)      return null;    else      return x.key;  }  /**   * 如果给定的键key小于二叉查找树的根节点的键，那么小于等于key的最大键一定出现在根节点的左子树中   * 如果给定的键key大于二叉查找树的根节点，那么只有当根节点右子树中存在大于等于key的节点时，   * 小于等于key的最大键才会出现在右子树中，否则根节点就是小于等于key的最大键   *    * @param x   * @param key   * @return   */  private Node floor(Node x, Key key) {    if (x == null)      return null;    int cmp = key.compareTo(x.key);    if (cmp == 0)      return x;    else if (cmp < 0)      return floor(x.left, key);    else {      Node t = floor(x.right, key);      if (t == null)        return x;      else        return t;    }  }  /**   * 向上取整：找出大于等于该键的最小键   *    * @param key   * @return   */  public Key ceiling(Key key) {    Node x = ceiling(root, key);    if (x == null)      return null;    else      return x.key;  }  /**   * 如果给定的键key大于二叉查找树的根节点的键，那么大于等于key的最小键一定出现在根节点的右子树中   * 如果给定的键key小于二叉查找树的根节点，那么只有当根节点左子树中存在大于等于key的节点时，   * 大于等于key的最小键才会出现在左子树中，否则根节点就是大于等于key的最小键   *    * @param x   * @param key   * @return   */  private Node ceiling(Node x, Key key) {    if (x == null)      return null;    int cmp = key.compareTo(x.key);    if (cmp == 0)      return x;    else if (cmp > 0) {      return ceiling(x.right, key);    } else {      Node t = floor(x.left, key);      if (t == null)        return x;      else        return t;    }  }  /**   * 选择排名为k的节点   *    * @param k   * @return   */  public Key select(int k) {    return select(root, k).key;  }  private Node select(Node x, int k) {    if (x == null)      return null;    int t = size(x.left);    if (t > k)      return select(x.left, k);    else if (t < k)      return select(x.right, k - t - 1);// 根节点也要排除掉    else      return x;  }  /**   * 查找给定键值的排名   *    * @param key   * @return   */  public int rank(Key key) {    return rank(key, root);  }  private int rank(Key key, Node x) {    if (x == null)      return 0;    int cmp = key.compareTo(x.key);    if (cmp < 0)      return rank(key, x.left);    else if (cmp > 0)      return 1 + size(x.left) + rank(key, x.right);    else      return size(x.left);  }  /**   * 删除最小键值对   */  public void deleteMin(){    root = deleteMin(root);  }  /**   * 不断深入根节点的左子树直到遇见一个空链接，然后将指向该节点的链接指向该结点的右子树   * 此时已经没有任何链接指向要被删除的结点，因此它会被垃圾收集器清理掉   * @param x   * @return   */  private Node deleteMin(Node x){    if(x.left == null) return x.right;    x.left = deleteMin(x.left);    x.n = size(x.left)+size(x.right) + 1;    return x;  }    public void deleteMax(){    root = deleteMax(root);  }  private Node deleteMax(Node x){    if(x.right == null ) return x.left;    x.right = deleteMax(x.right);    x.n = size(x.left)+size(x.right) + 1;    return x;  }    public void delete(Key key){    root = delete(root,key);  }  private Node delete(Node x, Key key){    if(x == null) return null;    int cmp = key.compareTo(x.key);    if(cmp < 0) x.left = delete(x.left,key);    else if(cmp > 0) x.right = delete(x.right,key);    else{      if(x.right == null) return x.left;      if(x.left == null ) return x.right;      /**       * 如果被删除节点有两个子树，将被删除节点暂记为t       * 从t的右子树中选取最小的节点x，将这个节点x的左子树设为t的左子树       * 这个节点x的右子树设为t的右子树中删除了最小节点的子树，这样就成功替换了t的位置       */      Node t = x;      x = min(t.right);      x.left = t.left;      x.right = deleteMin(t.right);    }    x.n = size(x.left) + size(x.right) +1;    return x;  }    public void print(){    print(root);  }  private void print(Node x){    if(x == null ) return;    print(x.left);    StdOut.println(x.key);    print(x.right);  }    public Iterable<Key> keys(){    return keys(min(),max());  }  public Iterable<Key> keys(Key lo, Key hi){    Queue<Key> queue = new Queue<Key>();    keys(root, queue, lo, hi);    return queue;  }  private void keys(Node x, Queue<Key> queue, Key lo, Key hi){    if(x == null) return;    int cmplo = lo.compareTo(x.key);    int cmphi = lo.compareTo(x.key);    if(cmplo < 0 ) keys(x.left,queue,lo,hi);    if(cmplo <= 0 && cmphi >= 0) queue.enqueue(x.key);    if(cmphi > 0 ) keys(x.right,queue,lo,hi);  }}