1.. 图解2-3树维持绝对平衡的原理：
2.. 红黑树与2-3树是等价的
3.. 红黑树的特点
- 简要概括如下：
- 所有节点非黑即红；根节点为黑；NULL节点为黑；红节点孩子为黑；黑平衡
4.. 实现红黑树的业务逻辑
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import java.util.ArrayList; public class RBTree<K extends Comparable<K>, V> { private static final boolean RED = true; private static final boolean BLACK = false; private class Node{ public K key; public V value; public Node left, right; public boolean color; public Node(K key, V value){ this.key = key; this.value = value; left = null; right = null; color = RED; } } private Node root; private int size; public RBTree(){ root = null; size = 0; } public int getSize(){ return size; } public boolean isEmpty(){ return size == 0; } // 判断节点node的颜色 private boolean isRed(Node node){ if(node == null) return BLACK; return node.color; } // node x // / 左旋转 / // T1 x ---------> node T3 // / / // T2 T3 T1 T2 private Node leftRotate(Node node){ Node x = node.right; // 左旋转 node.right = x.left; x.left = node; x.color = node.color; node.color = RED; return x; } // node x // / 右旋转 / // x T2 -------> y node // / / // y T1 T1 T2 private Node rightRotate(Node node){ Node x = node.left; // 右旋转 node.left = x.right; x.right = node; x.color = node.color; node.color = RED; return x; } // 颜色翻转 private void flipColors(Node node){ node.color = RED; node.left.color = BLACK; node.right.color = BLACK; } // 向红黑树中添加新的元素(key, value) public void add(K key, V value){ root = add(root, key, value); root.color = BLACK; // 最终根节点为黑色节点 } // 向以node为根的红黑树中插入元素(key, value)，递归算法 // 返回插入新节点后红黑树的根 private Node add(Node node, K key, V value){ if(node == null){ size ++; return new Node(key, value); // 默认插入红色节点 } if(key.compareTo(node.key) < 0) node.left = add(node.left, key, value); else if(key.compareTo(node.key) > 0) node.right = add(node.right, key, value); else // key.compareTo(node.key) == 0 node.value = value; if (isRed(node.right) && !isRed(node.left)) node = leftRotate(node); if (isRed(node.left) && isRed(node.left.left)) node = rightRotate(node); if (isRed(node.left) && isRed(node.right)) flipColors(node); return node; } // 返回以node为根节点的二分搜索树中，key所在的节点 private Node getNode(Node node, K key){ if(node == null) return null; if(key.equals(node.key)) return node; else if(key.compareTo(node.key) < 0) return getNode(node.left, key); else // if(key.compareTo(node.key) > 0) return getNode(node.right, key); } public boolean contains(K key){ return getNode(root, key) != null; } public V get(K key){ Node node = getNode(root, key); return node == null ? null : node.value; } public void set(K key, V newValue){ Node node = getNode(root, key); if(node == null) throw new IllegalArgumentException(key + " doesn't exist!"); node.value = newValue; } // 返回以node为根的二分搜索树的最小值所在的节点 private Node minimum(Node node){ if(node.left == null) return node; return minimum(node.left); } // 删除掉以node为根的二分搜索树中的最小节点 // 返回删除节点后新的二分搜索树的根 private Node removeMin(Node node){ if(node.left == null){ Node rightNode = node.right; node.right = null; size --; return rightNode; } node.left = removeMin(node.left); return node; } // 从二分搜索树中删除键为key的节点 public V remove(K key){ Node node = getNode(root, key); if(node != null){ root = remove(root, key); return node.value; } return null; } private Node remove(Node node, K key){ if( node == null ) return null; if( key.compareTo(node.key) < 0 ){ node.left = remove(node.left , key); return node; } else if(key.compareTo(node.key) > 0 ){ node.right = remove(node.right, key); return node; } else{ // key.compareTo(node.key) == 0 // 待删除节点左子树为空的情况 if(node.left == null){ Node rightNode = node.right; node.right = null; size --; return rightNode; } // 待删除节点右子树为空的情况 if(node.right == null){ Node leftNode = node.left; node.left = null; size --; return leftNode; } // 待删除节点左右子树均不为空的情况 // 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点 // 用这个节点顶替待删除节点的位置 Node successor = minimum(node.right); successor.right = removeMin(node.right); successor.left = node.left; node.left = node.right = null; return successor; } } public static void main(String[] args){ System.out.println("Pride and Prejudice"); ArrayList<String> words = new ArrayList<>(); if(FileOperation.readFile("pride-and-prejudice.txt", words)) { System.out.println("Total words: " + words.size()); RBTree<String, Integer> map = new RBTree<>(); for (String word : words) { if (map.contains(word)) map.set(word, map.get(word) + 1); else map.add(word, 1); } System.out.println("Total different words: " + map.getSize()); System.out.println("Frequency of PRIDE: " + map.get("pride")); System.out.println("Frequency of PREJUDICE: " + map.get("prejudice")); } System.out.println(); } }
5.. 向红黑树中添加新元素之后需要进行维护的过程示意图
