Java二叉查找树
gonghr 人气:0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
:one: key:关键字的值
:two: value:关键字的存储信息
:three: left:左节点的引用
:four: 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
:one: 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); }
: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
为根的二叉查找树中插入关键词为key
的结点 - 在
t
中查找key
,在查找失败处插入
:one: 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; }
: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
在 t
中找到关键词为 key
的结点,分三种情况删除 key
1.若 key
是叶子节点,则直接删除
2.若 key
只有一棵子树,则子承父业
3.若 key
既有左子树也有右子树,则找到 key
的后继结点,替换 key
和后继节点的值,然后删除后继节点(后继节点只有一棵子树,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。
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; }
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 = 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、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; } }
:bug: 方法测试:
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)
先序遍历:(key: 1 , value: 5) (key: 5 , value: 1) (key: 2 , value: 1) (key: 3 , value: 1)
1 5
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