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手写完美的二叉排序树

手写完美的二叉排序树

作者: jqboooo | 来源:发表于2018-12-14 14:45 被阅读0次

    二叉排序树,也叫搜索树、查找树,是一种有顺序的二叉树,或者是一颗空树,或者是一颗。

    它有以下特点:

    1、若左子树不为空,那么左子树上面的所有节点的关键字值都比根节点的关键字值小
    2、若右子树不为空,那么右子树上面的所有节点的关键字值都比根节点的关键字值大
    3、左右子树都为二叉树
    4、没有重复值(这一点在实际中可以忽略)
    

    应用场景

    在大数据中,越来越广泛的频繁使用二叉树,解决各种大数据查询的问题。在jdk1.8+版本中的大量的数据结构,都换成树来处理,而且树在大数据里,性能是做好的。它的作用就是快速查找。时间复杂度介于O(log₂n)到O(n)之间,数据量越大,越接近于O(log₂n),性能越优。

    代码实践

    1、使用孩子双亲法表示法来定义树

    public class Node<T extends Comparable> {
        T data;
        Node<T> leftChild;
        Node<T> rightChild;
        Node<T> parent;
    
        public Node(T data) {
            this.data = data;
            leftChild = null;
            rightChild = null;
            parent = null;
        }
    }
    

    2、首先实现新增的方法,在树的叶子节点添加一个节点

    /**
     * 新增一个节点
     *
     * @param data
     * @return
     */
    public Node put(T data) {
        Node<T> newNode = new Node(data);
        if (root == null) {
            root = newNode;
            return newNode;
        }
        Node<T> parent = null;
        Node<T> node = root;
        while (node != null) {
            parent = node;
            if (node.data.compareTo(data) > 0) {//小于跟节点就往左查询
                node = node.leftChild;
            } else if (node.data.compareTo(data) < 0) {//大于跟节点就往右查询
                node = node.rightChild;
            } else {//是重复值 就不理会了
                return node;
            }
        }
        if (parent.data.compareTo(data) > 0) {
            parent.leftChild = newNode;
        } else {
            parent.rightChild = newNode;
        }
        newNode.parent = parent;
        size++;
        return newNode;
    }
    

    3、查找节点

    public Node get(T data) {
        if (root == null) {
            return null;
        }
        Node<T> node = root;
        while (node != null) {
            if (node.data.compareTo(data) > 0) {
                node = node.leftChild;
            } else if (node.data.compareTo(data) < 0) {
                node = node.rightChild;
            } else {
                return node;
            }
        }
        return null;
    }   
    

    4、遍历树,采用中序遍历。

    /**
     * 中序遍历
     *
     * @param node
     */
    public void middleOrderTraseval(Node<T> node) {
        if (node == null) {
            return;
        }
        middleOrderTraseval(node.leftChild);
        System.out.print(node.data + "    ");
        middleOrderTraseval(node.rightChild);
    }
    

    5、删除节点(难点)分四种情况

    1. 要删除node是叶子节点

    if (left == null && right == null) {
            if (parent == null) {
                root = null;
            } else {
                if (parent.leftChild == node) {
                    parent.leftChild = null;
                } else {
                    parent.rightChild = null;
                }
                node.parent = null;
            }
        }
    

    2. 要删除node节点只有左节点

    else if (left != null && right == null) {
        if (parent == null) {
            root = left;
        } else {
            if (parent.leftChild == node) {
                parent.leftChild = left;
            } else {
                parent.rightChild = left;
            }
        }
        left.parent = parent;
        node.leftChild = null;
        node.parent = null;
    }
    

    3. 要删除node节点只有右节点

    else if (left == null && right != null) {
        if (parent == null) {
                root = right;
        } else {
            if (parent.leftChild == node) {
                    parent.leftChild = right;
            } else {
                    parent.rightChild = right;
                }
            }
            right.parent = parent;
            node.rightChild = null;
            node.parent = null;
    }
    

    4. 要删除node节点左右节点都有

    else if (left != null && right != null) {
        Node<T> leftMinNode = getLeftMinNode(right);
        // 1. 把右节点最小的节点接上node的左节点
        leftMinNode.leftChild = left;
        left.parent = leftMinNode;
        // 2. 如果右节点最小的节点,如果有右节点,则把右节点接上父节点
        Node<T> leftMinNodeParent = leftMinNode.parent;
        if (leftMinNode.rightChild != null) {
            if (leftMinNodeParent != node) {
                leftMinNodeParent.leftChild = leftMinNode.rightChild;
                leftMinNode.rightChild.parent = leftMinNodeParent;
            }
        } else {
            //没有右节点,则要把最小节点的父节点的左节点赋空
            leftMinNodeParent.leftChild = null;
        }
        // 3. 把右节点最小节点接上node的右节点上
        if (leftMinNode != right) {
            leftMinNode.rightChild = right;
        }
        right.parent = leftMinNode;
        // 4. 接上node的父节点
        if (parent == null) {
            root = leftMinNode;
        } else {
            if (parent.leftChild == node) {
                parent.leftChild = leftMinNode;
            } else {
                parent.rightChild = leftMinNode;
            }
        }
        leftMinNode.parent = parent;
        node.leftChild = null;
        node.rightChild = null;
        node.parent = null;
    }
    

    6、完整代码

    public class SearchBinaryTree<T extends Comparable> {
    
    //二叉树跟节点
    private Node<T> root;
    //二叉树大小
    private int size;
    
    /**
     * 新增一个节点
     *
     * @param data
     * @return
     */
    public Node put(T data) {
        Node<T> newNode = new Node(data);
        if (root == null) {
            root = newNode;
            return newNode;
        }
        Node<T> parent = null;
        Node<T> node = root;
        while (node != null) {
            parent = node;
            if (node.data.compareTo(data) > 0) {//小于跟节点就往左查询
                node = node.leftChild;
            } else if (node.data.compareTo(data) < 0) {//大于跟节点就往右查询
                node = node.rightChild;
            } else {//是重复值 就不理会了
                return node;
            }
        }
        if (parent.data.compareTo(data) > 0) {
            parent.leftChild = newNode;
        } else {
            parent.rightChild = newNode;
        }
        newNode.parent = parent;
    
        size++;
    
        return newNode;
    }
    
    public int getSize() {
        return size;
    }
    
    public Node get(T data) {
        if (root == null) {
            return null;
        }
        Node<T> node = root;
        while (node != null) {
            if (node.data.compareTo(data) > 0) {
                node = node.leftChild;
            } else if (node.data.compareTo(data) < 0) {
                node = node.rightChild;
            } else {
                return node;
            }
        }
        return null;
    }
    
    public void middleOrderTraseval() {
        middleOrderTraseval(root);
    }
    
    /**
     * 中序遍历
     *
     * @param node
     */
    public void middleOrderTraseval(Node<T> node) {
        if (node == null) {
            return;
        }
        middleOrderTraseval(node.leftChild);
        System.out.print(node.data + "    ");
        middleOrderTraseval(node.rightChild);
    }
    
    public void deleteNode(Node<T> node) {
        if (node == null) {
            return;
        }
        Node<T> left = node.leftChild;
        Node<T> right = node.rightChild;
        Node<T> parent = node.parent;
        //第一种情况:node是叶子节点
        if (left == null && right == null) {
            if (parent == null) {
                root = null;
            } else {
                if (parent.leftChild == node) {
                    parent.leftChild = null;
                } else {
                    parent.rightChild = null;
                }
                node.parent = null;
            }
        }
        //第二种情况:node只有左节点,没有右节点的情况
        else if (left != null && right == null) {
            if (parent == null) {
                root = left;
            } else {
                if (parent.leftChild == node) {
                    parent.leftChild = left;
                } else {
                    parent.rightChild = left;
                }
            }
            left.parent = parent;
            node.leftChild = null;
            node.parent = null;
        }
        //第三种情况:node只有右节点,没有左节点的情况
        else if (left == null && right != null) {
            if (parent == null) {
                root = right;
            } else {
                if (parent.leftChild == node) {
                    parent.leftChild = right;
                } else {
                    parent.rightChild = right;
                }
            }
            right.parent = parent;
            node.rightChild = null;
            node.parent = null;
        }
        //第四种情况:node左右节点都有的情况
        else if (left != null && right != null) {
            Node<T> leftMinNode = getLeftMinNode(right);
            // 1. 把右节点最小的节点接上node的左节点
            leftMinNode.leftChild = left;
            left.parent = leftMinNode;
            // 2. 如果右节点最小的节点,如果有右节点,则把右节点接上父节点
            Node<T> leftMinNodeParent = leftMinNode.parent;
            if (leftMinNode.rightChild != null) {
                if (leftMinNodeParent != node) {
                    leftMinNodeParent.leftChild = leftMinNode.rightChild;
                    leftMinNode.rightChild.parent = leftMinNodeParent;
                }
            } else {
                //没有右节点,则要把最小节点的父节点的左节点赋空
                leftMinNodeParent.leftChild = null;
            }
            // 3. 把右节点最小节点接上node的右节点上
            if (leftMinNode != right) {
                leftMinNode.rightChild = right;
            }
            right.parent = leftMinNode;
            // 4. 接上node的父节点
            if (parent == null) {
                root = leftMinNode;
            } else {
                if (parent.leftChild == node) {
                    parent.leftChild = leftMinNode;
                } else {
                    parent.rightChild = leftMinNode;
                }
            }
            leftMinNode.parent = parent;
            node.leftChild = null;
            node.rightChild = null;
            node.parent = null;
        }
        size--;
    }
    
    /**
     * 获取当前节点的左边最小的值,也就是最左边的节点
     *
     * @param node
     * @return
     */
    public Node<T> getLeftMinNode(Node<T> node) {
        if (node == null) {
            return null;
        }
        Node currentRoot = node;
        while (currentRoot.leftChild != null) {
            currentRoot = currentRoot.leftChild;
        }
        return currentRoot;
    }
    
    /**
     * 使用孩子双亲法表示法来定义树(实际双向链表)
     * @param <T> 对象
     */
    public class Node<T extends Comparable> {
        T data;
        Node<T> leftChild;
        Node<T> rightChild;
        Node<T> parent;
    
        public Node(T data) {
            this.data = data;
            leftChild = null;
            rightChild = null;
            parent = null;
        }
    }
    

    }

    7、测试

     @Test
    public void testBinarySortTree(){
        BinarySortTree<Integer> tree = new BinarySortTree();
        //5  2  7  3  4  1  6
        int[] array=new int[]{5,2,7,3,4,1,8,6,9};
        for (int i = 0; i < array.length; i++) {
            int i1 = array[i];
            tree.put(i1);
        }
        tree.middleOrderTraseval();
    
        System.out.println();
    
         //删除单个
        tree.deleteNode(tree.get(7));
    
        tree.middleOrderTraseval();
    
        System.out.println();
    
        for (int i : array) {
            tree.middleOrderTraseval();
            System.out.println("------------------------------");
            tree.deleteNode(tree.get(i));
        }
    
        System.out.println();
        for (int i = 0; i < array.length; i++) {
            tree.put(array[i]);
        }
        tree.middleOrderTraseval();
    
    }
    

    8、测试结果

    1    2    3    4    5    6    7    8    9    
    1    2    3    4    5    6    8    9    
    1    2    3    4    5    6    8    9    ------------------------------
    1    2    3    4    6    8    9    ------------------------------
    1    3    4    6    8    9    ------------------------------
    1    3    4    6    8    9    ------------------------------
    1    4    6    8    9    ------------------------------
    1    6    8    9    ------------------------------
    6    8    9    ------------------------------
    6    9    ------------------------------
    9    ------------------------------
    
    1    2    3    4    5    6    7    8    9        
    

    9、结束

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