紅黑樹

時間 2019-11-08

原文原文鏈接

二叉搜索樹基本操做均可以在O(h)內完成，然而數的高度較高時，可能並不比在鏈表上執行的快。紅黑樹是平衡搜索樹中的一種保證在最壞狀況下時間複雜度爲O(lg(n))html

紅黑樹的性質spa

每一個節點或是紅色或是黑色
根節點是黑色的
每一個葉節點是黑色的
若是一個節點是紅色，則兩個子節點是黑色的
每一個節點，到後代葉子節點的路徑上，包含相同數目的黑色節點

由於葉子節點是null，沒有辦法存儲顏色，因此用一個哨兵nil，其餘屬性p、left、right、key是任意值(null)爲好。code

由紅黑樹的性質可知，h最大爲2lg(n+1)htm

旋轉操做blog

圖片來源，其中關於紅黑樹的介紹更詳細圖片

LeftRotate(T, h)
{
    x = h.right;
    //x的左子樹成爲h的右子樹
    h.right = x.left;
    if (x.left != T.nil)
        x.left.parent = h;
    //x的parent是h的parent
    x.parent = h.parent;
    //h的parent對應h的子樹，成爲x
    if (h.parent == T.nil)
        T.root = x;
    else if (h == h.parent.left)
        h.parent.left = x;
    else
        h.parent.right = x;
    //將x的左子樹設置爲h
    x.left = h;
    h.parent = y;
}

RightRotate(T, x)
{
    h = x.left;
    //h的右子樹成爲x的左子樹
    x.left = h.right;
    if (h.right != T.nil)
        h.right.parent = x;
    //h的parent是x的parent
    h.parent = x.parent;
    //x的parent對應x的子樹，成爲h
    if (x.parent == T.nil)
        T.root = h;
    else if (x == x.parent.right)
        x.parent.right = h;
    else
        x.parent.left = h;
    //將h的右子樹設置爲x
    h.right = x;
    x.parent = y;
}

插入操做get

插入操做與搜索二叉樹操做基本相同。須要注意的是class

null被nil代替
插入的元素的子樹置爲nil
對插入的元素設置紅色
對插入的元素進行RbInsertFixup，保證符合紅黑樹性質

TreeInsert(T, z)
{
    y = T.nil;
    x = T.root;
    while (x != T.nil)
    {
        y = x;
        if (z.key < x.key)
            x = x.left;
        else
            x = x.right;
    }
    x.parent = y;
    if (y == T.nil)
        T.root = z;
    else if (z.key < y.key)
        y.left = z;
    else
        y.right = z;
    z.left = z.right = T.nil;
    z.color = RED;
    RbInsertFixup(T, z);
}

理解RbInsertFixup二叉樹

插入z，因爲z和父節點z.p都是red，違反了紅黑樹的性質。因爲z的叔節點y是red，對應於狀況1。節點從新上色，z上升一位，成爲了第二幅圖
z和父節點再次違規，z的叔節點是黑色的。由於z是z父節點的右孩子，對應於狀況2。執行一次左旋，獲得第三幅圖
z是父節點的左孩子，從新上色，執行右旋獲得第四幅圖
成爲合法的紅黑樹

RbInserFixup(T, z)
{
    while (z.parent.color == RED)
    {
        if (z.parent == z.parent.parent.left)
        {
            y = z.parent.parent.right;
            if (y.color == RED)
            {
                z.parent.color = BLACK;
                y.color = BLACK;
                z.parent.parent.color = RED;
                z = z.parent.parent;
            }
            else if (z == z.parent.right)
            {
                z = z.parent;
                LeftRotate(T, z);
            }
            z.parent.color = BLACK;
            z.parent.parent.color = RED;
        }
        else//將傷醫狀況的right與left換一下
        {
            y = z.parent.parent.left;
            if (y.color == RED)
            {
                z.parent.color = BLACK;
                y.color = BLACK;
                z.parent.parent.color = RED;
                z = z.parent.parent;
            }
            else if (z == z.parent.left)
            {
                z = z.parent;
                LeftRotate(T, z);
            }
            z.parent.color = BLACK;
            z.parent.parent.color = RED;
        }
    }
}

刪除操做搜索

狀況與二叉搜索樹類比。首先須要一個Transplant應用到紅黑樹上

RbTransplant(T, u, v)
{
    if (u.parent == T.nil)
        T.root = v;
    else if (u == u.parent.left)
        u.parent.left = v;
    else
        u.parent.right = v;
    v.parent = u.parent;
}

而後執行刪除

RbTreeDelete(T, z)
{
    y = z;
    yOriginalColor = y.color;
    if (z.left == T.nil)
    {
        x = z.right;
        RbTransplant(T, z, z.right);
    }
    else if (z.right == T.nil)
    {
        x = z.left;
        RbTransplant(T, z, z.left);
    }
    else
    {
        y = TreeMinimum(z.right);//z有雙子，則後繼爲右側最小
        yOriginalColor = y.color;
        x = y.right;
        if (y.parent == z)
        {
            x.parent = y;
        }
        else
        {
            BrTransplant(T, y, y.right);
            y.right = z.right;
            y.right.parent = y;
        }
        BrTransplant(T, z, y);
        y.left = z.left;
        y.left.p = y;
        y.color = z.color;
    }
    if (yOriginalColor == BLACK)
        RbDeleteFixup(T, x);
}

RbDeleteFixup(T, x)
{
    while (x != T.root && x.color == BLACK)
    {
        if (x == x.parent.left)
        {
            w = x.parent.right;
            if (w.color == RED)
            {
                w.color = BLACK;
                x.p.color = RED;
                LeftRotate(T, x.parent);
                w = x.parent.right;
            }
            if (w.left.color == BLACK && w.right.color == BLACK)
            {
                w.color = RED;
                x = x.parent;
            }
            else
            {
                w.left.color = BLACK;
                w.color = RED;
                RightRotate(T, w);
                w.x.parent.right;
            }
            w.color = x.parent.color;
            x.parent.color = BLACK;
            w.right.color = BLACK;
            LeftRotate(T, x.p);
            x = T.root;
        }
        else //change right and left
        {
            w = x.parent.left;
            if (w.color == RED)
            {
                w.color = BLACK;
                x.p.color = RED;
                LeftRotate(T, x.parent);
                w = x.parent.left;
            }
            if (w.right.color == BLACK && w.left.color == BLACK)
            {
                w.color = RED;
                x = x.parent;
            }
            else
            {
                w.right.color = BLACK;
                w.color = RED;
                RightRotate(T, w);
                w.x.parent.left;
            }
            w.color = x.parent.color;
            x.parent.color = BLACK;
            w.left.color = BLACK;
            LeftRotate(T, x.p);
            x = T.root;
        }
    }
    x.color = BLACK;
}

對紅黑樹的Extent

::動態順序統計

在紅黑樹的每一個節點中加入一個size屬性，表示這棵樹的節點數，其中哨兵的節點數size爲0。則

x.size=x.left.size+z.right.size+1

相關標籤/搜索

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每一个你不满意的现在，都有一个你没有努力的曾经。