【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
![](https://img2022.cnblogs.com/blog/2288178/202203/2288178-20220326095740068-1739757300.png)
节点结构
![](https://img2022.cnblogs.com/blog/2288178/202203/2288178-20220326173746464-876810742.png)
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 - 在
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 -
在
t
中找到关键词为key
的结点,分三种情况删除key
-
若
key
是叶子节点,则直接删除
-
若
key
只有一棵子树,则子承父业
-
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(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);
}
![](https://img2022.cnblogs.com/blog/2288178/202203/2288178-20220326180419880-687839535.png)
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)
1 5