Java數(shù)據(jù)結(jié)構(gòu)之二叉查找樹(shù)的實(shí)現(xiàn)

更新時(shí)間：2022年03月28日 15:49:13 作者：gonghr

二叉查找樹(shù)（亦稱二叉搜索樹(shù)、二叉排序樹(shù)）是一棵二叉樹(shù)，且各結(jié)點(diǎn)關(guān)鍵詞互異，其中根序列按其關(guān)鍵詞遞增排列。本文將通過(guò)示例詳細(xì)講解二叉查找樹(shù)，感興趣的可以了解一下

定義

二叉查找樹(shù)（亦稱二叉搜索樹(shù)、二叉排序樹(shù)）是一棵二叉樹(shù)，且各結(jié)點(diǎn)關(guān)鍵詞互異，其中根序列按其關(guān)鍵詞遞增排列。

等價(jià)描述：二叉查找樹(shù)中任一結(jié)點(diǎn) P，其左子樹(shù)中結(jié)點(diǎn)的關(guān)鍵詞都小于 P 的關(guān)鍵詞，右子樹(shù)中結(jié)點(diǎn)的關(guān)鍵詞都大于 P 的關(guān)鍵詞，且結(jié)點(diǎn) P 的左右子樹(shù)也都是二叉查找樹(shù)

節(jié)點(diǎn)結(jié)構(gòu)

:one: key：關(guān)鍵字的值

:two: value：關(guān)鍵字的存儲(chǔ)信息

:three: left：左節(jié)點(diǎn)的引用

:four: right：右節(jié)點(diǎn)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;
}

為了代碼簡(jiǎn)潔，本文不考慮屬性的封裝，一律設(shè)為 public

查找算法

思想：利用二叉查找樹(shù)的特性，左子樹(shù)值小于根節(jié)點(diǎn)值，右子樹(shù)值大于根節(jié)點(diǎn)值，從根節(jié)點(diǎn)開(kāi)始搜索

如果目標(biāo)值等于某節(jié)點(diǎn)值，返回該節(jié)點(diǎn)
如果目標(biāo)值小于某節(jié)點(diǎn)值，搜索該節(jié)點(diǎn)的左子樹(shù)
如果目標(biāo)值大于某節(jié)點(diǎn)值，搜索該節(jié)點(diǎn)的右子樹(shù)

:one: 遞歸版本

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);
    }

:two: 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 為根的二叉查找樹(shù)中插入關(guān)鍵詞為 key 的結(jié)點(diǎn)
在 t 中查找 key ，在查找失敗處插入

:one: 遞歸版本

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;  // 如果二叉查找樹(shù)中已經(jīng)存在關(guān)鍵字，則替換該結(jié)點(diǎn)的值
        }
        return t;
    }

:two: 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;    // 如果二叉查找樹(shù)中已經(jīng)存在關(guān)鍵字，則替換該結(jié)點(diǎn)的值
                return;
            }
        }
        if(key.compareTo(pre.key) < 0) {
            pre.left = new BSTNode<>(key, value);
        } else {
            pre.right = new BSTNode<>(key, value);
        }
    }

刪除算法

在以 t 為根的二叉查找樹(shù)中刪除關(guān)鍵詞值為 key 的結(jié)點(diǎn)

在 t 中找到關(guān)鍵詞為 key 的結(jié)點(diǎn)，分三種情況刪除 key

1.若 key 是葉子節(jié)點(diǎn)，則直接刪除

2.若 key 只有一棵子樹(shù)，則子承父業(yè)

3.若 key 既有左子樹(shù)也有右子樹(shù)，則找到 key 的后繼結(jié)點(diǎn)，替換 key 和后繼節(jié)點(diǎn)的值，然后刪除后繼節(jié)點(diǎn)（后繼節(jié)點(diǎn)只有一棵子樹(shù)，轉(zhuǎn)化為第二種情況）。

后繼結(jié)點(diǎn)是當(dāng)前結(jié)點(diǎn)的右子樹(shù)的最左結(jié)點(diǎn)，如果右子樹(shù)沒(méi)有左子樹(shù)，則后繼節(jié)點(diǎn)就是右子樹(shù)的根節(jié)點(diǎn)。

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，遞歸處理右子樹(shù)
        else if(t.key.compareTo(key) > 0) t.left = removeByRecursion(t.left, key);   // key小，遞歸處理左子樹(shù)
        else {
            if(t.right == null) return t.left;  // 情況一、二一起處理
            if(t.left == null) return t.right;  // 情況一、二一起處理
            BSTNode<K, V> node = t.right;       // 情況三：右子樹(shù)沒(méi)有左子樹(shù)
            if (node.left == null) {
                node.left = t.left;
            } else {                            // 情況三：右子樹(shù)有左子樹(shù)
                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;
    }

完整代碼

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("后序遍歷：");
        postorder(root);
        System.out.println();
    }

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

    public void inorderTraverse() {
        System.out.print("中序遍歷：");
        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 = 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;
    }

    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;
        }
        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;
            if (key.compareTo(p.key) < 0) p = p.left;
            else if (key.compareTo(p.key) > 0) p = p.right;
            else {
                p.value = value;    // 如果二叉查找樹(shù)中已經(jīng)存在關(guān)鍵字，則替換該結(jié)點(diǎn)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，遞歸處理右子樹(shù)
        else if(t.key.compareTo(key) > 0) t.left = removeByRecursion(t.left, key);   // key小，遞歸處理左子樹(shù)
        else {
            if(t.right == null) return t.left;  // 情況一、二一起處理
            if(t.left == null) return t.right;  // 情況一、二一起處理
            BSTNode<K, V> node = t.right;       // 情況三：右子樹(shù)沒(méi)有左子樹(shù)
            if (node.left == null) {
                node.left = t.left;
            } else {                            // 情況三：右子樹(shù)有左子樹(shù)
                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;
    }
}

:bug: 方法測(cè)試：

public class Main {
    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.inorderTraverse();
        tree.postorderTraverse();
        tree.preorderTraverse();
        BSTNode<Integer, Integer> node = tree.searchByIteration(1);
        System.out.println(node.key + " " + node.value);
    }
}

中序遍歷：(key: 1 , value: 5) (key: 2 , value: 1) (key: 3 , value: 1) (key: 5 , value: 1)
后序遍歷：(key: 3 , value: 1) (key: 2 , value: 1) (key: 5 , value: 1) (key: 1 , value: 5)
先序遍歷：(key: 1 , value: 5) (key: 5 , value: 1) (key: 2 , value: 1) (key: 3 , value: 1)
1 5

以上就是Java數(shù)據(jù)結(jié)構(gòu)之二叉查找樹(shù)的實(shí)現(xiàn)的詳細(xì)內(nèi)容，更多關(guān)于Java二叉查找樹(shù)的資料請(qǐng)關(guān)注腳本之家其它相關(guān)文章！

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