二叉树

树

@(数据结构)

[TOC]

树的定义

（递归）一棵树是一些节点的集合。这个集合可以是空集；若不是空集，则树由称作根的节点 r 以及 0 个或多个非空的（子）树 $T_1，T_2，···，T_k$ 组成，这些子树中每一棵的根都被来自根 r 的一条有向边所连结。

树的实现

//树节点的声明
class TreeNode
{
    Object element;
    TreeNode firstChild;
    TreeNode netSibling;
}

将每个节点的所有儿子都放到树节点的链表中。

树的遍历

• 先序遍历
• 后序遍历
• 中序遍历

二叉树

二叉树（binary tree）是一棵树，其中每个节点都不能有多于两个的儿子。

二叉树平均深度为 $O(\sqrt{N})$，最大深度为 $N$。
二叉查找树的平均深度为 $O(log N)$。

//二叉树节点类
class BinaryNode
{
    //Friendly data;accessible by other package toutines
    Object element;//The data in the node
    BinaryNode left;//Left child
    BinaryNode right;//right child
}

查找树ADT——二叉查找树

使二叉树成为查找树的性质是，对于树中的每个节点 X ，它的左子树中所有项的值小于 X 中的项，而它的右子树中所有项的值大于 X 中的项。

//BinaryNode类
private static class BinaryNode<AnyType>
{
    //Constructors
    BinaryNode(AnyType theElement)
    {this(theElement, null, null);}

    BinaryNode(AnyType theElement, BinaryNode<AnyType> lt, BinaryNode<AnyType> rt)
    {element = theElement; left = lt; right = rt;}

    AnyType element;//The data in the node
    BinaryNode<AnyType> left;//Left child
    BinaryNode<AnyType> right;//Right child
}

二叉查找树架构

//二叉查找树架构
public class BinarySearchTree<AnyType extends comparable<? super AnyType>>
{
    private static class BinaryNode<AnyType>
    {
        //Constructors
        BinaryNode(AnyType theElement)
        {this(theElement, null, null);}

        BinaryNode(AnyType theElement, BinaryNode<AnyType> lt, BinaryNode<AnyType> rt)
        {element = theElement; left = lt; right = rt;}

        AnyType element;//The data in the node
        BinaryNode<AnyType> left;//Left child
        BinaryNode<AnyType> right;//Right child
    }

    private BinaryNode<AnyType> root;

    public BinarySearchTree()
    { root = null; }

    public void makeEmpty()
    { root = null; }
    public boolean isEmpty()
    { return root == null; }

    public boolean contains( AnyType x )
    { return contains( x, root ); }
    public AnyType findMin()
    {
        if (isEmpty()) throw new UnderflowException();
        return findMin(root).element;
    }
    public AnyType finMax()
    {
        if (isEmpty()) throw new UnderflowException();
        return finMax(roow).element;
    }
    public void insert(AnyType x)
    { root = insert(x,root); }
    public void remove(AnyType x)
    { root = remove(x,root); } 
    public void printTree()
    {
        if (isEmpty())
            System.out.println("Empty tree");
        else
            printTree(root);
    }

    private boolean contains(AnyType x, BinaryNode<AnyType> t)
    {
        if (t == null) 
            return false;
        int compareResult = x.compareTo(t.element);

        if(compareResult < 0)
            return contains(x, t.left);
        else if(compareResult > 0)
            return contains(x, t.right);
        else
            return true; //Match
    }
    private BinaryNode<AnyType> findMin(BinaryNode<AnyType> t)
    {
        if(t == null)
            return null;
        else if(t.left == null)
            return t;
        return findMin(t.left);
    }
    private BinaryNode<AnyType> finMax(BinaryNode<AnyType> t)
    {
        if(t != null)
            while(t.right != null)
                t = t.right;

        return t;
    }

    private BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> t)
    {
        if(t == null)
            return new BinaryNode<>(x, null, null);

        int compareResult = x.compareTo(t.element);

        if(compareResult < 0)
            t.left = insert(x, t.left);
        else if(compareResult > 0)
            t.right = insert(x, t.right);
        else
            ;//Duplicate; do nothing
        return t;
    }
    private BinaryNode<AnyType> remove(AnyType x, BinaryNode<AnyType> t)
    {
        if(t == null)
            return t;//Item not found; do nothing

        int compareResult = x.compareTo(t.element);

        if(compareResult < 0)
            t.left = remove(x, t.left);
        else if(compareResult > 0)
            t.right = remove(x, t.right);
        else if(t.left != null && t.right != null)//Two children
        {
            t.element = findMin(t.right).element;
            t.right = remove(t.element, t.right);
        }
        else
            t = (t.left != null) ? t.left : t.right;
        return t;
    }
    private void printTree(BinaryNode<AnyType> t)
    {
        if (t != null) {
            printTree(t.left);
            System.out.println(t.element);
            printTree(t.right);
        }
    }


}

contains方法

如果树 $T$ 中含有项 $X$ 的节点，那么这个操作需要返回true，如果这样的节点不存在则返回false。树的结构使这种操作很简单。如果 $T$ 是空集，那么久返回false。否则，如果存储在 $T$ 处的项是 $X$ ，那么可以返回true。否则，我们对数 $T$ 的左子树或右子树进行一次递归调用，则依赖于 $X$ 与存储在 $T$ 中的项的关系。

/**
 * Internal method to find an item in a subtree
 * @param  x is item to search for.
 * @param  t the node that roots the subtree.
 * @return true if the item is found; false otherwise.
 */

//二叉查找树的contains操作
private boolean contains(AnyType x, BinaryNode<AnyType> t)
    {
        if (t == null) 
            return false;
        int compareResult = x.compareTo(t.element);

        if(compareResult < 0)
            return contains(x, t.left);
        else if(compareResult > 0)
            return contains(x, t.right);
        else
            return true; //Match
    }

    //递归用while循环代替
    while(compareResult <0)
    {
        t=t.left;
        compareResult = x.compareTo(t.element);
    }

算法表达式的简明性是以速度的降低为代价的。

findMin方法和findMax方法

这两个方法分别返回树中包含最小元和最大元的节点的引用。为执行findMin，从根开始并且只要有左儿子就向左进行。 终止点就是最小的元素。findMax除分支朝向右儿子其余过程相同。

//用递归编写findMin，用非递归编写findMax
/**
* Internal method to find the smallest item in a subtree
* @param  t the node that roots the subtree.
* @return node containing the smallest item
*/
private BinaryNode<AnyType> findMin(BinaryNode<AnyType> t)
{
    if(t == null)
        return null;
    else if(t.left == null)
        return t;
    return findMin(t.left);
}
/**
* Internal method to find the largest item in a subtree
* @param  t the node that roots the subtree.
* @return node containing the largest item.
*/
private BinaryNode<AnyType> finMax(BinaryNode<AnyType> t)
{
    if(t != null)
        while(t.right != null)
            t = t.right;

    return t;
    }

insert方法

/**
 * Internal method to insert into a subtree
 * @param  x the item to insert
 * @param  t the node that roots the subtree
 * @return the new root of the subtree
 */ 
private BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> t)
{
    if(t == null)
        return new BinaryNode<>(x, null, null);

    int compareResult = x.compareTo(t.element);

    if(compareResult < 0)
        t.left = insert(x, t.left);
    else if(compareResult > 0)
        t.right = insert(x, t.right);
    else
        ;//Duplicate; do nothing
    return t;
}

remove方法

    /**
     * Internal method to remove from a subtree
     * @param  x the item to remove.
     * @param  t the node that roots the subtree.
     * @return the new root of the subtree
     */
    private BinaryNode<AnyType> remove(AnyType x, BinaryNode<AnyType> t)
    {
        if(t == null)
            return t;//Item not found; do nothing

        int compareResult = x.compareTo(t.element);

        if(compareResult < 0)
            t.left = remove(x, t.left);
        else if(compareResult > 0)
            t.right = remove(x, t.right);
        else if(t.left != null && t.right != null)//Node that has two children
        {
            t.element = findMin(t.right).element;//Find the minimum item of right subtree
            t.right = remove(t.element, t.right);//Remove the node of minimum item recursively            
        }
        else
            t = (t.left != null) ? t.left : t.right;//Node that has one children; parent of the node roots subtree of the node
        return t;
    }

• 如果节点是树叶，可以直接删除。
• 如果节点有一个儿子，这该节点需要在其父节点调整自己的链以绕过该节点
• 如果节点有两个儿子，一般的删除策略是用其右子树的最小的数据代替该节点，并在右子树中递归地删除那个最小的节点

另外，如果删除的次数不多，通常使用的策略是懒惰删除（lazy deletion）：当一个元素要被删除时，它仍留在树中，而只是被标记为删除。

AVL树

AVL树是带有平衡条件的二叉查找树。
这个平衡条件必须要容易保持，而且它保证树的深度须是 $O(log N)$ 。
一个AVL树是其每个节点的左子树和右子树的高度最多差 1 的二叉查找树（空树的高度定义为 -1）。

可以知道，在高度为 $h$ 的AVL树中，最少节点数 $S(h)=S(h-1)+S(h-2)+1$ 给出。
对于 $h=0, S(h)=1; h=1, S(h)=2$ 。
函数 $S(h)$ 与斐波那契数密切相关。

那么重点来了，对于AVL树的插入操作，有可能破坏树的平衡性。这时候，我们就需要在这一步插入完成之前恢复平衡的性质。

可以知道，从插入的节点往上，逆行到根，若发生平衡信息改变，那么改变的节点一定在这条路径上。我们需要找出这个需要重新平衡的节点 $\alpha$ 。

对于节点 $\alpha$ ，不平衡条件可能出现在一下四种操作中：

1. 对 $\alpha$ 的左儿子的左子树进行一次插入（LL）。
2. 对 $\alpha$ 的左儿子的右子树进行一次插入（LR）。
3. 对 $\alpha$ 的右儿子的左子树进行一次插入（RL）。
4. 对 $\alpha$ 的右儿子的右子树进行一次插入（RR）。

对于1和4，是插入发生在外边的情况，通过对树的一次单旋转而完成调整。对于2和3，是插入发生在内部的情况，通过对树的一次双旋转而完成调整。

这里先对AvlNode类进行定义：

private static class AvlNode<AnyType>
{
    //Constructors
    AvlNode(AnyType theElement)
    {this(theElement, null, null);}

    AvlNode(AnyType theElement, AvlNode<AnyType> lt, AvlNode<AnyType> rt)
    {element = theElement; left = lt; right = rt; height = 0;}

    AnyType element;//The data in the code
    AvlNode<AnyType> left;//Left child
    AvlNode<AnyType> right;//Right child
    int height;//Height
}

然后需要一个返回节点高度的方法：

    //返回AVL树的节点高度
    /**
     * return the height of node t, or -1, if null.
     */
    private int height(AvlNode<AnyType> t)
    {
        return t == null ? -1 : t.height;
    }

单旋转

LL单旋转

/**
 * Rotate binary tree node with left child.
 * For AVL trees, this is a single rotation for case 1.
 * Update heights, then return new root.
 */
private AvlNode<AnyType> RotationWithLeftChild(AvlNode<AnyType> k2) 
{  
    AVLTreeNode<AnyType> k1 = k2.left;  

    k2.left = k1.right;  
    k1.right = k2;  

    k2.height = Math.max( height(k2.left), height(k2.right)) + 1;  
    k1.height = Math.max( height(k1.left), k2.height) + 1;  

    return k1;  
}

RR单旋转

/**
 * Rotate binary tree node with right child.
 * For AVL trees, this is a single rotation for case 4.
 * Update heights, then return new root.
 */
private AvlNode<AnyType> RotationWithRightChild(AvlNode<AnyType> k1) 
{  
    AVLTreeNode<AnyType> k2 = k1.right;  

      k1.right = k2.left;  
       k2.left = k1;  

       k1.height = Math.max( height(k1.left), height(k1.right)) + 1;  
    k1.height = Math.max( height(k2.right), k1.height) + 1;  

    return k2;  
}

双旋转

LR双旋转

    /**
     * Double rotate binary tree node: first left child
     * with its right child; then node k3 with new left child.
     * For AVL trees, this is a double rotation for case 2.
     * Update heights, then return new root.
     */
    private AvlNode<AnyType> doubleWithLeftChild(AvlNode<AnyType> k3)
    {
        k3.left = RotationWithRightChild(k3.left);
        return RotationWithLeftChild(k3);
    }

RL双旋转

    /**
     * Double rotate binary tree node: first right child
     * with its left child; then node k1 with new right child.
     * For AVL trees, this is a double rotation for case 3.
     * Update heights, then return new root.
     */
    private AvlNode<AnyType> doubleWithRightChild(AvlNode<AnyType> k1)
    {
        k1.right = RotationWithRightChild(k1.right);
        return RotationWithLeftChild(k1);
    }

AVL树的插入方法

插入方法就是前文中的insert方法，只是在最后一行调用平衡的方法以保持AVL树的平衡性。
```java
/**

 * Internal method to insert into a subtree.
 * @param  x the item to insert.
 * @param  t the node that roots the subtree.
 * @return the new root of the subtree.
 */
private AvlNode<AnyType> insert(AnyType x, AvlNode<AnyType> t)
{
    if(t == null)
        return new    AvlNode<>(x, null, null);

    int compareResult = x.compareTo(t.element);

    if(compareResult < 0)
        t.left = insert(x, t.left);
    else if(compareResult > 0)
        t.right = insert(x, t.right);
    else
        ;//Duplicate; do nothing
    return balance(t);
}

private static final int ALLOWED_IMBALLANCE = 1;

//Assume t is either balanced of within one of being balanced
private AvlNode<AnyType> balance(AvlNode<AnyType> t)
{
    if(t == null)
        return t;

    if(height(t.left) - height(t.right) > ALLOWED_IMBALLANCE)
        if(height(t.left.left) >= height(t.left.right))
            t = RotationWithLeftChild(t);
        else
            t = doubleWithLeftChild(t);
    else
    if(height(t.right) - height(t.left) > ALLOWED_IMBALLANCE)
        if(height(t.right.right) >= height(t.right.left))
            t = RotationWithRightChild(t);
        else
            t = doubleWithRightChild(t);

    t.height = Math.max(height(t.left), height(t.right)) + 1;
    return t;
}

###AVL树的删除方法
> 和AVL树的插入一样，只用在前文的删除方法最后加上一行调用平衡的方法即可。

private AvlNode<AnyType> remove(AnyType x, AvlNode<AnyType> t)
{
    if(t == null)
        return t;//Item not found; do nothing

    int compareResult = x.compareTo(t.element);

    if(compareResult < 0)
        t.left = remove(x, t.left);
    else if(compareResult > 0)
        t.right = remove(x, t.right);
    else if(t.left != null && t.right != null)//Node that has two children
    {
        t.element = findMin(t.right).element;//Find the minimum item of right subtree
        t.right = remove(t.element, t.right);//Remove the node of minimum item recursively            
    }
    else
        t = (t.left != null) ? t.left : t.right;//Node that has one children; parent of the node roots subtree of the node
    return balance(t);
}

```