【数据结构与算法】手撕二叉查找树

二叉查找树

定义

  • 二叉查找树(亦称二叉搜索树、二叉排序树)是一棵二叉树,且各结点关键词互异,其中根序列按其关键词递增排列。

  • 等价描述:二叉查找树中任一结点 P,其左子树中结点的关键词都小于 P 的关键词,右子树中结点的关键词都大于 P 的关键词,且结点 P 的左右子树也都是二叉查找树

节点结构

1️⃣ key:关键字的值
2️⃣ value:关键字的存储信息
3️⃣ left:左节点的引用
4️⃣ right:右节点的引用

class BSTNode<K extends Comparable<K>,V>{
    public K key;
    public V value;

    public BSTNode<K,V> left;
    public BSTNode<K,V> right;
}

为了代码简洁,本文不考虑属性的封装,一律设为 public

查找算法

思想:利用二叉查找树的特性,左子树值小于根节点值,右子树值大于根节点值,从根节点开始搜索

  • 如果目标值等于某节点值,返回该节点
  • 如果目标值小于某节点值,搜索该节点的左子树
  • 如果目标值大于某节点值,搜索该节点的右子树

1️⃣ 递归版本

    public BSTNode<K, V> searchByRecursion(K key) {
        return searchByRecursion(root, key);
    }

    private BSTNode<K, V> searchByRecursion(BSTNode<K, V> t, K key) {
        if (t == null || t.key == key) return t;
        else if (key.compareTo(t.key) < 0) return searchByRecursion(t.left, key);
        else return searchByRecursion(t.right, key);
    }

2️⃣ 迭代版本

    public BSTNode<K,V> searchByIteration(K key) {
        BSTNode<K,V> p = this.root;
        while(p != null) {
            if(key.compareTo(p.key) < 0) p = p.left;
            else if(key.compareTo(p.key) > 0) p = p.right;
            else return p;
        }
        return null;
    }

插入算法

  • 在以 t 为根的二叉查找树中插入关键词为 key 的结点
  • t 中查找 key,在查找失败处插入

1️⃣ 递归版本

public void insertByRecursion(K key, V value) {
        this.root = insertByRecursion(root, key, value);
    }

    private BSTNode<K, V> insertByRecursion(BSTNode<K, V> t, K key, V value) {
        if (t == null) {
            return new BSTNode<>(key, value);
        } 
        else if (key.compareTo(t.key) < 0) t.left = insertByRecursion(t.left, key, value);
        else if (key.compareTo(t.key) > 0) t.right = insertByRecursion(t.right, key, value);
        else {
            t.value = value;  // 如果二叉查找树中已经存在关键字,则替换该结点的值
        }
        return t;
    }

2️⃣ 迭代版本

    public void insertByIteration(K key, V value) {
        BSTNode<K, V> p = root;
        if (p == null) {
            root = new BSTNode<>(key, value);
            return;
        }
        BSTNode<K, V> pre = null;
        while (p != null) {
            pre = p;
            if (key.compareTo(p.key) < 0) p = p.left;
            else if (key.compareTo(p.key) > 0) p = p.right;
            else {
                p.value = value;    // 如果二叉查找树中已经存在关键字,则替换该结点的值
                return;
            }
        }
        if(key.compareTo(pre.key) < 0) {
            pre.left = new BSTNode<>(key, value);
        } else {
            pre.right = new BSTNode<>(key, value);
        }
    }

删除算法

  • 在以 t 为根的二叉查找树中删除关键词值为 key 的结点

  • t 中找到关键词为 key 的结点,分三种情况删除 key

    • key 是叶子节点,则直接删除

    • key 只有一棵子树,则子承父业

    • key 既有左子树也有右子树,则找到 key 的后继结点,替换 key 和后继(前驱)节点的值,然后删除后继节点(后继/前驱节点只有一棵子树,转化为第二种情况)。
      后继结点是当前结点的右子树的最左结点,如果右子树没有左子树,则后继节点就是右子树的根节点。
      前驱结点时当前节点的左子树的最右结点,如果左子树没有右子树,则前驱结点就是左子树的根节点。

1️⃣ 写法一

    public void removeByRecursion(K key) {
        this.root = removeByRecursion(root, key);
    }
    private BSTNode<K, V> removeByRecursion(BSTNode<K, V> t, K key) {
        if(t == null) return root;
        else if(t.key.compareTo(key) < 0) t.right = removeByRecursion(t.right, key); // key大,递归处理右子树
        else if(t.key.compareTo(key) > 0) t.left = removeByRecursion(t.left, key);   // key小,递归处理左子树
        else {
            if(t.right == null) return t.left;  // 情况一、二一起处理
            if(t.left == null) return t.right;  // 情况一、二一起处理
            BSTNode<K, V> node = t.right;       // 情况三:右子树没有左子树
            if (node.left == null) {
                node.left = t.left;
            } else {                            // 情况三:右子树有左子树
                BSTNode<K, V> pre = null;
                while (node.left != null) {
                    pre = node;
                    node = node.left;
                }
                t.key = node.key;
                t.value = node.value;
                pre.left = node.right;
            }
        }
        return t;
    }

2️⃣ 写法二

情况三也进行递归处理,写法更简单(考虑的前驱结点)

    private BSTNode<K, V> removeByRecursion(BSTNode<K, V> t, K key) {
        if (t == null) return root;
        else if (t.key.compareTo(key) < 0) t.right = removeByRecursion(t.right, key); // key大,递归处理右子树
        else if (t.key.compareTo(key) > 0) t.left = removeByRecursion(t.left, key);   // key小,递归处理左子树
        else {
            else if (t.right == null) return t.left;
            else {
                BSTNode<K, V> pre = t.left;
                while (pre.right != null) {
                    pre = pre.right;
                }
                t.key = pre.key;
                t.value = pre.value;
                t.left = removeByRecursion(t.left, t.key);
            }
        }
        return t;
    }

完整代码

class BSTNode<K extends Comparable<K>, V> {
    public K key;
    public V value;

    public BSTNode<K, V> left;
    public BSTNode<K, V> right;

    public BSTNode(K key, V value) {
        this.key = key;
        this.value = value;
    }
}

class BSTree<K extends Comparable<K>, V> {
    public BSTNode<K, V> root;

    private void inorder(BSTNode<K, V> root) {
        if (root != null) {
            inorder(root.left);
            System.out.print("(key: " + root.key + " , value: " + root.value + ") ");
            inorder(root.right);
        }
    }

    private void preorder(BSTNode<K, V> root) {
        if (root != null) {
            System.out.print("(key: " + root.key + " , value: " + root.value + ") ");
            preorder(root.left);
            preorder(root.right);
        }
    }

    private void postorder(BSTNode<K, V> root) {
        if (root != null) {
            postorder(root.left);
            postorder(root.right);
            System.out.print("(key: " + root.key + " , value: " + root.value + ") ");
        }
    }

    public void postorderTraverse() {
        System.out.print("后序遍历:");
        postorder(root);
        System.out.println();
    }

    public void preorderTraverse() {
        System.out.print("先序遍历:");
        preorder(root);
        System.out.println();
    }

    public void inorderTraverse() {
        System.out.print("中序遍历:");
        inorder(root);
        System.out.println();
    }

    public BSTNode<K, V> searchByRecursion(K key) {
        return searchByRecursion(root, key);
    }

    private BSTNode<K, V> searchByRecursion(BSTNode<K, V> t, K key) {
        if (t == null || t.key == key) return t;
        else if (key.compareTo(t.key) < 0) return searchByRecursion(t.left, key);
        else return searchByRecursion(t.right, key);
    }

    public BSTNode<K, V> searchByIteration(K key) {
        BSTNode<K, V> p = root;
        int cmp = key.compareTo(p.key);
        while (p != null) {
            if (cmp < 0) p = p.left;
            else if (cmp > 0) p = p.right;
            else return p;
        }
        return null;
    }

    public void insertByRecursion(K key, V value) {
        this.root = insertByRecursion(root, key, value);
    }

    private BSTNode<K, V> insertByRecursion(BSTNode<K, V> t, K key, V value) {
        if (t == null) {
            return new BSTNode<>(key, value);
        }
        int cmp = key.compareTo(t.key);
        if (cmp < 0) t.left = insertByRecursion(t.left, key, value);
        else if (cmp > 0) t.right = insertByRecursion(t.right, key, value);
        else {
            t.value = value;
        }
        return t;
    }

    public void insertByIteration(K key, V value) {
        BSTNode<K, V> p = root;
        if (p == null) {
            root = new BSTNode<>(key, value);
            return;
        }
        BSTNode<K, V> pre = null;
        while (p != null) {
            pre = p;
            int cmp = key.compareTo(p.key);
            if (cmp < 0) p = p.left;
            else if (cmp > 0) p = p.right;
            else {
                p.value = value;    // 如果二叉查找树中已经存在关键字,则替换该结点的值
                return;
            }
        }
        if (key.compareTo(pre.key) < 0) {
            pre.left = new BSTNode<>(key, value);
        } else {
            pre.right = new BSTNode<>(key, value);
        }
    }

    public void removeByRecursion(K key) {
        this.root = removeByRecursion(root, key);
    }

    private BSTNode<K, V> removeByRecursion(BSTNode<K, V> t, K key) {
        if (t == null) return root;
        else if (t.key.compareTo(key) < 0) t.right = removeByRecursion(t.right, key); // key大,递归处理右子树
        else if (t.key.compareTo(key) > 0) t.left = removeByRecursion(t.left, key);   // key小,递归处理左子树
        else {
//            if (t.right == null) return t.left;  // 情况一
//            if (t.left == null) return t.right;  // 情况二
//            BSTNode<K, V> node = t.right;       // 情况三:右子树没有左子树
//            if (node.left == null) {
//                node.left = t.left;
//            } else {                            // 情况三:右子树有左子树
//                BSTNode<K, V> pre = null;
//                while (node.left != null) {
//                    pre = node;
//                    node = node.left;
//                }
//                t.key = node.key;
//                t.value = node.value;
//                pre.left = node.right;
            if (t.left == null) return t.right;
            else if (t.right == null) return t.left;
            else {
                BSTNode<K, V> pre = t.left;
                while (pre.right != null) {
                    pre = pre.right;
                }
                t.key = pre.key;
                t.value = pre.value;
                t.left = removeByRecursion(t.left, t.key);
            }
        }
        return t;
    }
}

🐛 方法测试:

    public static void main(String[] args) {
        BSTree<Integer, Integer> tree = new BSTree<>();
//        tree.insertByRecursion(1, 1);
//        tree.insertByRecursion(5, 1);
//        tree.insertByRecursion(2, 1);
//        tree.insertByRecursion(4, 1);
//        tree.insertByRecursion(3, 1);
//        tree.insertByRecursion(1, 2);
        tree.insertByIteration(1, 1);
        tree.insertByIteration(5, 1);
        tree.insertByIteration(2, 1);
        tree.insertByIteration(4, 1);
        tree.insertByIteration(3, 1);
        //tree.insertByIteration(1, 5);
        tree.removeByRecursion(4);
        tree.preorderTraverse();
        tree.inorderTraverse();
        tree.postorderTraverse();
        BSTNode<Integer, Integer> node = tree.searchByIteration(1);
        System.out.println(node.key + " " + node.value);
    }
先序遍历:(key: 1 , value: 5) (key: 5 , value: 1) (key: 2 , value: 1) (key: 3 , value: 1) 
中序遍历:(key: 1 , value: 5) (key: 2 , value: 1) (key: 3 , value: 1) (key: 5 , value: 1) 
后序遍历:(key: 3 , value: 1) (key: 2 , value: 1) (key: 5 , value: 1) (key: 1 , value: 5) 
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posted @ 2022-03-27 21:42  gonghr  阅读(201)  评论(0编辑  收藏  举报