Tree

BST

Hibbard deletion.

2-3

2-3 Tree is Balance Tree:

插入:

• 1+1=2node -> 3node
• 1+2=3node -> 4node -> 2node
• 将 4node 结点中间元素移至父结点, 其余 2 元素分离为子 2node 节点

Red-Black

• 基于 2-3 tree, 将 3node 用红色标记.
• 关键: 将红色标记向上传递至根部.

基本性质:

1. 非红即黑.
2. 根黑.
3. 叶黑: T.null 为黑哨兵.
4. 红父孩子黑.
5. 简单路径同黑.
6. 右孩子不红:
   • Flip: 父黑两孩红 -> 父红两孩黑.
   • Left rotate: 父黑右红 -> 父左旋变红, 右孩子变黑.

基本操作:

1. 插入: 插入红点, 旋转 + 重新着色 (反色) 保持红黑性质.
2. 删除: 删除红点, 旋转 + 重新着色 (反色) 保持红黑性质.

    // is node x red; false if x is null ?
    private boolean isRed(Node x) {
        if (x == null) return false;
        return x.color == RED;
    }

    // make a right-leaning link lean to the left
    private Node rotateLeft(Node h) {
        // assert (h != null) && isRed(h.right);
        Node x = h.right;

        h.right = x.left;
        x.left = h;

        x.color = x.left.color;
        x.left.color = RED;

        x.size = h.size;
        h.size = size(h.left) + size(h.right) + 1;

        return x;
    }

    // make a left-leaning link lean to the right
    private Node rotateRight(Node h) {
        // assert (h != null) && isRed(h.left);
        Node x = h.left;

        h.left = x.right;
        x.right = h;

        x.color = x.right.color;
        x.right.color = RED;

        x.size = h.size;
        h.size = size(h.left) + size(h.right) + 1;

        return x;
    }

    // flip the colors of a node and its two children
    private void flipColors(Node h) {
        // h must have opposite color of its two children
        // assert (h != null) && (h.left != null) && (h.right != null);
        // assert (!isRed(h) &&  isRed(h.left) &&  isRed(h.right))
        //    || (isRed(h)  && !isRed(h.left) && !isRed(h.right));
        h.color = !h.color;
        h.left.color = !h.left.color;
        h.right.color = !h.right.color;
    }

    // insert the key-value pair in the subtree rooted at h
    private Node put(Node h, Key key, Value val) {
        // insert/put new node as left/right child of leaf node
        if (h == null) return new Node(key, val, RED, 1);

        int cmp = key.compareTo(h.key);
        if      (cmp < 0) h.left  = put(h.left,  key, val);
        else if (cmp > 0) h.right = put(h.right, key, val);
        else              h.val   = val;

        // fix-up any right-leaning links
        if (isRed(h.right) && !isRed(h.left))      h = rotateLeft(h);
        if (isRed(h.left)  &&  isRed(h.left.left)) h = rotateRight(h);
        if (isRed(h.left)  &&  isRed(h.right))     flipColors(h);

        h.size = size(h.left) + size(h.right) + 1;

        return h;
    }

    public void put(Key key, Value val) {
        if (key == null) {
            throw new IllegalArgumentException("first argument to put() is null");
        }

        if (val == null) {
            delete(key);
            return;
        }

        root = put(root, key, val);
        root.color = BLACK;
        // assert check();
    }

    // Assuming that h is red and both h.left and h.left.left
    // are black, make h.left or one of its children red.
    private Node moveRedLeft(Node h) {
        // assert (h != null);
        // assert isRed(h) && !isRed(h.left) && !isRed(h.left.left);

        flipColors(h);
        if (isRed(h.right.left)) {
            h.right = rotateRight(h.right);
            h = rotateLeft(h);
            flipColors(h);
        }
        return h;
    }

    // Assuming that h is red and both h.right and h.right.left
    // are black, make h.right or one of its children red.
    private Node moveRedRight(Node h) {
        // assert (h != null);
        // assert isRed(h) && !isRed(h.right) && !isRed(h.right.left);
        flipColors(h);
        if (isRed(h.left.left)) {
            h = rotateRight(h);
            flipColors(h);
        }
        return h;
    }

    // restore red-black tree invariant
    private Node balance(Node h) {
        // assert (h != null);

        if (isRed(h.right))                      h = rotateLeft(h);
        if (isRed(h.left) && isRed(h.left.left)) h = rotateRight(h);
        if (isRed(h.left) && isRed(h.right))     flipColors(h);

        h.size = size(h.left) + size(h.right) + 1;
        return h;
    }

    // delete the key-value pair with the minimum key rooted at h
    private Node deleteMin(Node h) {
        if (h.left == null)
            return null;

        if (!isRed(h.left) && !isRed(h.left.left))
            h = moveRedLeft(h);

        h.left = deleteMin(h.left);
        return balance(h);
    }

    /**
     * Removes the smallest key and associated value from the symbol table.
     * @throws NoSuchElementException if the symbol table is empty
     */
    public void deleteMin() {
        if (isEmpty()) throw new NoSuchElementException("BST underflow");

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = deleteMin(root);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

    // delete the key-value pair with the maximum key rooted at h
    private Node deleteMax(Node h) {
        if (isRed(h.left))
            h = rotateRight(h);

        if (h.right == null)
            return null;

        if (!isRed(h.right) && !isRed(h.right.left))
            h = moveRedRight(h);

        h.right = deleteMax(h.right);

        return balance(h);
    }

    /**
     * Removes the largest key and associated value from the symbol table.
     * @throws NoSuchElementException if the symbol table is empty
     */
    public void deleteMax() {
        if (isEmpty()) throw new NoSuchElementException("BST underflow");

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = deleteMax(root);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

    // delete the key-value pair with the given key rooted at h
    private Node delete(Node h, Key key) {
        // assert get(h, key) != null;

        if (key.compareTo(h.key) < 0)  {
            if (!isRed(h.left) && !isRed(h.left.left))
                h = moveRedLeft(h);
            h.left = delete(h.left, key);
        }
        else {
            if (isRed(h.left))
                h = rotateRight(h);
            if (key.compareTo(h.key) == 0 && (h.right == null))
                return null;
            if (!isRed(h.right) && !isRed(h.right.left))
                h = moveRedRight(h);
            if (key.compareTo(h.key) == 0) {
                Node x = min(h.right);
                h.key = x.key;
                h.val = x.val;
                // h.val = get(h.right, min(h.right).key);
                // h.key = min(h.right).key;
                h.right = deleteMin(h.right);
            }
            else h.right = delete(h.right, key);
        }
        return balance(h);
    }

    /**
     * Removes the specified key and its associated value from this symbol table
     * (if the key is in this symbol table).
     *
     * @param  key the key
     * @throws IllegalArgumentException if {@code key} is {@code null}
     */
    public void delete(Key key) {
        if (key == null) throw new IllegalArgumentException("argument to delete() is null");
        if (!contains(key)) return;

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = delete(root, key);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

B

• t: 每个内部结点至少 t 个孩子 (t-1 个 key), 至多 2t 个孩子 (2t-1 个 key).
• 插入/删除:
  • 下溯的同时,分裂满结点.
  • 合并/上溯的同时,合并空结点.

K-Dimensional

分隔空间数据: e.g. 左子树：左下方, 右子树：右上方.

Fibonacci Heap

BST + 循环双向链表:

• 一个根树 (根结点) 循环双向链表.
• n 个孩子循环双向链表: 每个根树的每层结点形成一个循环双向链表.

Edit Distance

给定 cost (edit operation) 时的最小编辑费用.