Question

我有一个二进制搜索树类，我想知道，使用高度变量，如何使用它来计算每个单独节点的路径长度？

例如：

＆＃34; E＆＃34;的路径大小（高度）是1，路径大小＆＃34; A＆＃34;是2，路径大小＆＃34; M＆＃34;是3，路径大小＆＃34; X＆＃34;是1.如何在使用put（）方法插入节点时执行此操作？

private Key key;           // sorted by key
private Value val;         // associated data
private Node left, right;  // left and right subtrees
private int N;             // number of nodes in subtree
private int height;

private Node put(Node x, Key key, Value val)
{
    if (x == null) { return new Node(key, val, 1); }
    x.height = -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.val   = val;
    }
    x.N = 1 + size(x.left) + size(x.right);
    return x;
}

Answer 1

根据定义，叶子的路径是其父级+ 1的路径。

您可以通过添加到您的方法轻松实现这一目标：

private Node put(Node x, Key key, Value val, int height){
 if (x == null) { return new Node(key, val, height); }
 .
 .
 .
}

首次来电应该是put(null,key,val,0)

只需按每次调用增加高度，例如：

x.left  = put(x.left,  key, val, ++height);

当然，您必须为Node构造函数添加高度。

Answer 2

您可以拥有如下所示的树/节点类关系。这摆脱了尴尬的x.N = 1 +大小（x.left）+大小（x.right）;因为单个节点的大小没有意义。

public class Tree<K extends Comparable<K>, V> {

  private Node head;
  private int size; // number of nodes in subtree
  private int height;

  public Node put(K key, V value) {
    size++;
    if (head == null) {
      head = new Node(key, value, 0);
      return head;
    }
    Node result = head.put(key, value);
    if (result.height > height) {
      height = result.height;
    }
    return result;
  }

  private class Node {

    private K key; // sorted by key
    private V value; // associated data
    private Node left, right; // left and right subtrees
    private int height;

    public Node(K key, V value, int height) {
      this.key = key;
      this.value = value;
      this.height = height;
    }

    private Node put(K key, V value) {
      int cmp = key.compareTo(this.key);
      if (cmp < 0) {
        if (left == null) {
          left = new Node(key, value, height + 1);
          return left;
        } else {
          return left.put(key, value);
        }
      } else if (cmp > 0) {
        if (right == null) {
          right = new Node(key, value, height + 1);
          return right;
        } else {
          return right.put(key, value);
        }
      } else {
        this.value = value;
        return this;
      }
    }
  }
}

为二进制搜索树中的每个节点添加路径长度

2 个答案: