二叉搜尋樹資料結構完整實現

文章首發於此，BST是後面自平衡二叉樹AVL樹，B樹等資料結構的基礎，所以理解BST的基本性質和操作很有必要，如果讀者對BST不是很瞭解可以查下wiki或者是參考嚴蔚敏的《資料結構與演算法》或者《演算法導論》，對於該資料結構有較詳細解釋，下面是我查詢資料實現的Java版本。

package com.mars.search;public class BinarySeachTree {    private Node mRoot;    private Node getRoot() {        return mRoot;    }    public int search(Node root, int key) {        if (root == null) {            return 0xffffffff;        }        if (key < root.key) {            return search(root.leftNode, key);        }        if (key > root.key) {            return search(root.rightNode, key);        } else {            return key;        }    }    public void insert(int data) {        if (mRoot == null) {            mRoot = new Node(data, null, null);            // mRoot.key = data;        }        Node root = mRoot;        while (root != null) {            if (data == root.key) {                return;            } else if (data < root.key) {                if (root.leftNode == null) {                    root.leftNode = new Node(data, null, null);                    return;                } else {                    root = root.leftNode;                }            } else {                if (root.rightNode == null) {                    root.rightNode = new Node(data, null, null);                    return;                } else {                    root = root.rightNode;                }            }        }    }    /*     * http://courses.csail.mit.edu/6.006/spring11/rec/rec03.pdf Description:     * Remove the node x from the binary search tree, making the necessary     * adjustments to the binary search tree to maintain its properties. (Note     * that this operation removes a specified node from the tree. If you wanted     * to delete a key k from the tree, you would have to first call find(k) to     * find the node with key k and then call delete to remove that node) Case 1:     * x has no children. Just delete it (i.e. change parent node so that it     * doesn’t point to x) Case 2: x has one child. Splice out x by linking x’s     * parent to x’s child Case 3: x has two children. Splice out x’s successor     * and replace x with x’s successor     *      * BST刪除分析可以看這裡 http://webdocs.cs.ualberta.ca/~holte/T26/del-from-bst.html     */    public boolean delete(int key) {        Node root = deleteNode(this.mRoot, key);        if (root != null) {            return true;        }        return false;    }    /**     *      * 刪除有左右子樹節點的時候其實等價於：1將右子樹的最左葉子節點的值覆蓋要刪除節點的值，     * 然後刪除右子樹最左葉子節點。 遞迴的過程中找到刪除節點然後將右子樹的最左子葉子節點的     * 值賦給刪除節點，然後刪除右子樹的最左葉子節點。這步在遞迴退出的時候就做了，如果想不     * 通可以細細的看下第一個if語句     *      * */    private Node deleteNode(Node root, int key) {        if (key < root.key) {            if (root.leftNode != null) {                root.leftNode = deleteNode(root.leftNode, key);            } else {                return null;            }        } else if (key > root.key) {            if (root.rightNode != null) {                root.rightNode = deleteNode(root.rightNode, key);            } else {                return null;            }        } else {            if (root.leftNode == null) {                root = root.rightNode;            } else if (root.rightNode == null) {                root = root.leftNode;            } else {                Node newRoot = findMin(root);                root.key = newRoot.key;                // 此步遞迴調用相當於把右子樹的最左子樹的左葉子給刪了，因為遞迴退出的時候leftNode = null;                //如果Node的結構裡有指向parent的引用，刪除會方便些                root.rightNode = deleteNode(root.rightNode, root.key);            }        }        return root;    }    // 左子樹的最右葉子節點    private Node findMin(Node root) {        Node newRoot = root.rightNode;        while (newRoot.leftNode != null) {            newRoot = newRoot.leftNode;        }        return newRoot;    }    public void travel(Node root) {        Node node = root;        if (node == null) {            return;        }        travel(node.leftNode);        System.out.print(node.key + " ");        travel(node.rightNode);    }    public static class Node {        int key;        Node leftNode;        Node rightNode;        Node(int key, Node left, Node right) {            this.key = key;            this.leftNode = left;            this.rightNode = right;        }    }    public static void main(String[] args) {        BinarySeachTree bst = new BinarySeachTree();        int[] a = {                0, 4, 3, 2, 5, 1, 7, 6, 8, 9        };        for (int i : a) {            bst.insert(i);        }        bst.travel(bst.getRoot());        int i = bst.search(bst.getRoot(), 3);        System.out.println("\nsearch :" + i);        System.out.println("after delete:");        bst.delete(4);        bst.travel(bst.getRoot());    }}

參考資料http://en.wikipedia.org/wiki/Binary_search_tree

資料結構與演算法