特点：结合有序组和链表的优点

查找

public Node find(int key) {
        Node current = root;
        while(current.iData != key) {
            if(key<current.iData)
                current = current.leftChild;
            else
                current = current.rightChild;
            if(current == null) {
                return null;
            }
        }
        return current;
    }

插入：顺着查找的思路，在返回前插入

public void insert(int id, double dd) {
        Node newNode = new Node();
        newNode.iData = id;
        newNode.dData = dd;
        if(root == null) {
            root = newNode;
        } else {
            Node current = root;
            Node parent;
            while(true) {
                parent = current;
                if(id < current.iData) {
                    current = current.leftChild;
                    if(current == null) {
                        parent.leftChild = newNode;
                        return;
                    }
                }else {
                    current = current.rightChild;
                    if(current == null) {
                        parent.rightChild = newNode;
                        return;
                    }
                }
            }
        }
    }

遍历树
中序遍历：
1.调用自身来遍历节点的左子树
2.访问自己节点
3….右子树

private void inOrder(node localRoot) {
  if(localRoot != null) {
      inOrder(localRoot.leftChild);
      System.out.println(localRoot.iData + "");
      inOrder(localRoot.rightChild);
    }
}  

前序：
1.访问自己节点
2.自身左子树
3.自身右子树
后序：自己想

最小值和最大值：最下最左左子树，最下最右右子树

删除节点
三种情况：
1.该节点是叶节点（没有子节点）
2.该节点有一个子节点
3.该节点有两个子节点

public boolean delete(int key) {
        Node current = root;
        Node parent = root;
        boolean isLeftChild = true;
        
        //和查找一样
        while(current.iData != key) {
            parent = current;
            if(key<current.iData) {
                isLeftChild = true;
                current = current.leftChild;
            }
            else {
                isLeftChild = false;
                current = current.rightChild;
            }
            if(current == null) {
                return null;
            }
        }
        
        //找到之后
        //第一种情况
        if(current.leftChild == null && current.rightChild == null) {
            if(current == root)
                root = null;
            else if(isLeftChild)
                parent.leftChild = null;
            else
                parent.rightChild = null;
        }
        //第二种情况：没有左节点或没有右节点，都要考虑被删除节点可能是根
        else if(current.rightChild == null) {
            if(current == root)
                root = current.leftChild;
            else if(isLeftChild)
                parent.leftChild = current.leftChild;
            else
                parent.rightChild = current.rightChild;
        }
        else if(current.leftChild == null) {
            if(current == root)
                root = current.rightChild;
            else if(isLeftChild)
                parent.leftChild = current.leftChild;
            else
                parent.rightChild = current.rightChild;
        }
        
        
        //第三种情况：双节点又分两种情况，第一种后继节点是delNode的右节点参考图8.19，第二种后继节点是delNode右子节点的左后代参考图8.20
        Node successor = getSuccessor(current);
        
        if(current == root) 
            root = successor;
        else if(isLeftChild)
            parent.leftChild = successor;
        else
            parent.rightChild = current.leftChild;
    }
    return true;
}

//查找后继节点，参照图8.17和8.18是两种情况
private node getSuccessor(node delNode) {
    Node SuccessorParent = delNode;
    Node successor = delNode;
    Node current = delNode.rightChild;
        //参照图8.17，38是需要删除的节点
    while(current != null) {
        SuccessorParent = successor;
        successor = current;
        current = current.leftChild;
    }
    
    //参照图8.18，假如不等于，这两句代码是实现第二种情况的树旋转
    if(successor != delNode.rightChild){
        SuccessorParent.leftChild = successor.rightChild;
        successor.rightChild = delNode.rightChild;
    }
    return successor;
}

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哈夫曼
专用于压缩文本

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重点解释：用10表示S,用00表示空格后，不能再用01和11，因为它们是其它字符的前缀，不够就用三位来组合：000，001，010，011，100，110，111. A是010，I是110，因为除去10和00开始的组合，在用四位表示其它字符，依此类推

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