第五周上：Balanced Search Trees

1. 2-3 search trees

1. 每个Node有1或2个key

   • 2-node：one key，two children
   • 3-node：two keys，three children

2. Perfect balance：

   • 每一条path，从root到null-link都是相同长度

3. Symmetric order

   • 横向从左到右升序

4. Search

   • search key与各个Node的key比较
   • 沿着对应的link（recursively），直到找到相同的key

5. Insertion

   • 若每个key都已经有2个key，将新key加入这个3-node形成临时的4-node

   • 将这个4-node中间值的key晋升为parent

   • 重复直到结束

   • 当这条search path从root起的每一个node都是2-node，那么再加入一个key，length+1

2. Red-Black BSTs

Represent 2-3 tree as a BST

1. Definition:

   • 每一个Node最多只有一个red-link
   • 每一条path从root到null-link都有相同数量的black-links
   • Red-link只在左侧
   • 当将red-link平方，red-black BST就是2-3 tree

2. Insertion

   1. 找到的null-link的parent node只有black-link

      • 若key小于parent node，带着red-link加在左边就完成
      • 若key大于parent node，带着red-link加在右边，再rotate left即可

   2. 找到的null-link的parent node已经有red-link（相当于在2-3tree中，3-node要临时变4-node）

      • case 1（larger：已有a，b(parent)，插入c）：找到null-link，加入新的node（key=c， red-link），然后flip color
      • case 2（smaller：已有b，c(parent)，插入a）：找到null-link，加入新的node（key=a， red-link）。现在b有两个red-links，不符合规定。c进行rotate right操作，回归case 1，然后flip color
      • case 3（middle：已有a，c(parent)，插入b）：找到null-link，加入新的node（key=b， red-link）。此时red-link在a的右侧，不符合规定。b进行rotate left，回归case2，然后c进行rotate right操作，最后flip color

3. Java implementation

   public class RBtree<Key extends Comparable<Key>, Value>{
   
    private static final boolean RED = true;
    private static final boolean BLACK = false;
    private Node root; // root of BST
   
    private class Node{
        Key key;
        Value val;
        Node left, right;
        boolean color;  // color of parent link
   
        public Node(Key key, Value val, boolean color){
            this.key = key;
            this.val = val;
            this.color = color;
        }
    }
   
    public void put(Key key, Value val){
         root = put(root, key, val);
       }
   
    private Node put(Node h, Key key, Value val){
        // find the null-link, insert at bottom (and color it red)
        if(h == null){ 
            return new Node(key, val, RED);
        }
   
        int cmp = key.compareTo(x.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;
        }
   
        // Right child red, left child black: rotate left.
        if(isRed(h.right) && !isRed(h.left)){
            h = rotateLeft(h);
        }
        // Left child, left-left grandchild red: rotate right.
        if(isRed(h.left) && isRed(h.left.left)){
            h = rotateRight(h);
        }
        // Both children red: flip colors.
        if(isRed(h.left) && isRed(h.right)){
            flipColor(h);
        }
   
        return h;
   
    }
   
    private boolean isRed(Node x){
        if(x == null){
            return false;
        }
        return x.color == RED;
    }
   
    // Orient a (temporarily) right-leaning red link to lean left.
    private Node rotateLeft(Node h){
        // assert isRed(h.right);
        Node x = h.right;
        h.right = x.left;
        x.left = h;
        x.color = h.color;
        h.color = RED;
        return x
    }
   
    // Orient a left-leaning red link to (temporarily) lean right.
    private Node rotateRight(Node h){
        // assert isRed(h.left);
        Node x = h.left;
        h.left = x.right;
        x.right = h;
        x.color = h.color;
        h.color = RED;
        return x;
   
    }
   
    // Recolor to split a (temporary) 4-node.
    private void flipColors(Node h){
        // assert !isRed(h);
        // assert isRed(h.left);
        // assert isRed(h.right);
        h.color = RED;
        h.left.color = BLACK;
        h.right.color = BLACK;
    }
   
    public Value get(Key key){
        Node x = root;
        while(x != null){
            int cmp = key.compareTo(x.key);
            if(cmp < 0){
                x = x.left;
            }else if(cmp > 0){
                x = x.right;
            }else{
                return null;
            }
        }
    } 
   }