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note/datastructureAlgorithm/book/datastructure/BinarySortTree.md
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## 二叉排序树(BST) | ||
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### 介绍 | ||
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二叉排序树:BST: (Binary Sort(Search) Tree), 对于二叉排序树的任何一个非叶子节点,要求左子节点的值比当前节点的值小,右子节点的值比当前节点的值大。 | ||
特别说明:如果有相同的值,可以将该节点放在左子节点或右子节点 | ||
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比如针对前面的数据 (7, 3, 10, 12, 5, 1, 9) ,对应的二叉排序树为: | ||
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![bst](img/bst/bst01.jpg) | ||
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### 二叉排序树创建和遍历 | ||
一个数组创建成对应的二叉排序树,并使用中序遍历二叉排序树,比如: 数组为 Array(7, 3, 10, 12, 5, 1, 9) , 创建成对应的二叉排序树为 : | ||
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```java | ||
7 | ||
/ \ | ||
3 10 | ||
/ \ / \ | ||
1 5 9 12 | ||
``` | ||
### 二叉排序树的删除 | ||
二叉排序树的删除情况比较复杂,有下面三种情况需要考虑 | ||
1) 删除叶子节点 (比如:2, 5, 9, 12) | ||
2) 删除只有一颗子树的节点 (比如:1) | ||
3) 删除有两颗子树的节点. (比如:7, 3,10 ) | ||
4) 操作的思路分析 | ||
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```java | ||
//对删除结点的各种情况的思路分析: | ||
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第一种情况: | ||
删除叶子节点 (比如:2, 5, 9, 12) | ||
思路 | ||
(1) 需求先去找到要删除的结点 targetNode | ||
(2) 找到 targetNode 的 父结点 parent | ||
(3) 确定 targetNode 是 parent 的左子结点 还是右子结点 | ||
(4) 根据前面的情况来对应删除左子结点 parent.left = null | ||
右子结点 parent.right = null; | ||
第二种情况: 删除只有一颗子树的节点 比如 1 | ||
思路 | ||
(1) 需求先去找到要删除的结点 targetNode | ||
(2) 找到 targetNode 的 父结点 parent | ||
(3) 确定 targetNode 的子结点是左子结点还是右子结点 | ||
(4) targetNode 是 parent 的左子结点还是右子结点 | ||
(5) 如果 targetNode 有左子结点 | ||
5. 1 如果 targetNode 是 parent 的左子结点 | ||
parent.left = targetNode.left; | ||
5.2 如果 targetNode 是 parent 的右子结点 | ||
parent.right = targetNode.left; | ||
(6) 如果 targetNode 有右子结点 | ||
6.1 如果 targetNode 是 parent 的左子结点 | ||
parent.left = targetNode.right; | ||
6.2 如果 targetNode 是 parent 的右子结点 | ||
parent.right = targetNode.right | ||
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情况三 : 删除有两颗子树的节点. (比如:7, 3,10 ) | ||
思路 | ||
(1) 需求先去找到要删除的结点 targetNode | ||
(2) 找到 targetNode 的 父结点 parent | ||
(3) 从 targetNode 的右子树找到最小的结点 | ||
(4) 用一个临时变量,将 最小结点的值保存 temp = 11 | ||
(5) 删除该最小结点 | ||
(6) targetNode.value = temp | ||
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``` | ||
### 代码实现 | ||
```java | ||
package com.javayh.advanced.datastructure.tree.sort; | ||
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/** | ||
* <p> | ||
* 二叉排序树 | ||
* </p> | ||
* | ||
* @author Dylan | ||
* @version 1.0.0 | ||
* @since 2020-12-31 8:04 PM | ||
*/ | ||
public class BinarySortTreeDemo { | ||
public static void main(String[] args) { | ||
int[] arr = {7, 3, 10, 12, 5, 1, 9, 2}; | ||
BinarySortTree binarySortTree = new BinarySortTree(); | ||
//循环的添加结点到二叉排序树 | ||
for (int value : arr) { | ||
binarySortTree.add(new Node(value)); | ||
} | ||
//中序遍历二叉排序树 | ||
System.out.println("中序遍历二叉排序树~"); | ||
binarySortTree.infixOrder(); // 1, 3, 5, 7, 9, 10, 12 | ||
//测试一下删除叶子结点 | ||
binarySortTree.delNode(12); | ||
binarySortTree.delNode(5); | ||
binarySortTree.delNode(10); | ||
binarySortTree.delNode(2); | ||
binarySortTree.delNode(3); | ||
binarySortTree.delNode(9); | ||
System.out.println("del后中序遍历二叉排序树~"); | ||
binarySortTree.infixOrder(); | ||
} | ||
} | ||
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class BinarySortTree { | ||
private Node root; | ||
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//查找要删除的结点 | ||
public Node search(int value) { | ||
if (root == null) { | ||
return null; | ||
} else { | ||
return root.search(value); | ||
} | ||
} | ||
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//查找父结点 | ||
public Node searchParent(int value) { | ||
if (root == null) { | ||
return null; | ||
} else { | ||
return root.searchParent(value); | ||
} | ||
} | ||
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//编写方法: | ||
//1. 返回的 以 node 为根结点的二叉排序树的最小结点的值 | ||
//2. 删除 node 为根结点的二叉排序树的最小结点 | ||
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/** | ||
* @param node 传入的结点(当做二叉排序树的根结点) | ||
* @return 返回的 以 node 为根结点的二叉排序树的最小结点的值 | ||
*/ | ||
public int delRightTreeMin(Node node) { | ||
Node target = node; | ||
//循环的查找左子节点,就会找到最小值 | ||
while (target.left != null) { | ||
target = target.left; | ||
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} | ||
//这时 target 就指向了最小结点 | ||
//删除最小结点 | ||
delNode(target.value); | ||
return target.value; | ||
} | ||
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//删除结点 | ||
public void delNode(int value) { | ||
if (root == null) { | ||
return; | ||
} else { | ||
//1.需求先去找到要删除的结点 targetNode | ||
Node targetNode = search(value); | ||
//如果没有找到要删除的结点 | ||
if (targetNode == null) { | ||
return; | ||
} | ||
//如果我们发现当前这颗二叉排序树只有一个结点 | ||
if (root.left == null && root.right == null) { | ||
root = null; | ||
return; | ||
} | ||
//去找到 targetNode 的父结点 | ||
Node parent = searchParent(value); | ||
//如果要删除的结点是叶子结点 | ||
if (targetNode.left == null && targetNode.right == null) { | ||
//判断 targetNode 是父结点的左子结点,还是右子结点 | ||
if (parent.left != null && parent.left.value == value) { //是左子结点 | ||
parent.left = null; | ||
} else if (parent.right != null && parent.right.value == value) {//是由子结点 | ||
parent.right = null; | ||
} | ||
} else if (targetNode.left != null && targetNode.right != null) { //删除有两颗子树的节点 | ||
int minVal = delRightTreeMin(targetNode.right); | ||
targetNode.value = minVal; | ||
} else { // 删除只有一颗子树的结点 | ||
//如果要删除的结点有左子结点 | ||
if (targetNode.left != null) { | ||
if (parent != null) { | ||
//如果 targetNode 是 parent 的左子结点 | ||
if (parent.left.value == value) { | ||
parent.left = targetNode.left; | ||
} else { // targetNode 是 parent 的右子结点 | ||
parent.right = targetNode.left; | ||
} | ||
} else { | ||
root = targetNode.left; | ||
} | ||
} else { //如果要删除的结点有右子结点 | ||
if (parent != null) { | ||
//如果 targetNode 是 parent 的左子结点 | ||
if (parent.left.value == value) { | ||
parent.left = targetNode.right; | ||
} else { //如果 targetNode 是 parent 的右子结点 | ||
parent.right = targetNode.right; | ||
} | ||
} else { | ||
root = targetNode.right; | ||
} | ||
} | ||
} | ||
} | ||
} | ||
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//添加结点的方法 | ||
public void add(Node node) { | ||
if (root == null) { | ||
root = node;//如果 root 为空则直接让 root 指向 node | ||
} else { | ||
root.add(node); | ||
} | ||
} | ||
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//中序遍历 | ||
public void infixOrder() { | ||
if (root != null) { | ||
root.infixOrder(); | ||
} else { | ||
System.out.println("二叉排序树为空,不能遍历"); | ||
} | ||
} | ||
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} | ||
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class Node { | ||
int value; | ||
Node left; | ||
Node right; | ||
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public Node(int value) { | ||
this.value = value; | ||
} | ||
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/** | ||
* 查找要删除的结点 | ||
* | ||
* @param value 希望删除的结点的值 | ||
* @return 如果找到返回该结点,否则返回 null | ||
*/ | ||
public Node search(int value) { | ||
//找到就是该结点 | ||
if (value == this.value) { | ||
return this; | ||
}//如果查找的值小于当前结点,向左子树递归查找 | ||
else if (value < this.value) { | ||
//如果左子结点为空 | ||
if (this.left == null) { | ||
return null; | ||
} | ||
return this.left.search(value); | ||
} else { //如果查找的值不小于当前结点,向右子树递归查找 | ||
if (this.right == null) { | ||
return null; | ||
} | ||
return this.right.search(value); | ||
} | ||
} | ||
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public Node searchParent(int value) { | ||
//如果当前结点就是要删除的结点的父结点,就返回 | ||
if ((this.left != null && this.left.value == value) || | ||
(this.right != null && this.right.value == value)) { | ||
return this; | ||
} else { | ||
//如果查找的值小于当前结点的值, 并且当前结点的左子结点不为空 | ||
if (value < this.value && this.left != null) { | ||
//向左子树递归查找 | ||
return this.left.searchParent(value); | ||
} else if (value >= this.value && this.right != null) { | ||
//向右子树递归查找 | ||
return this.right.searchParent(value); | ||
} else { | ||
// 没有找到父结点 | ||
return null; | ||
} | ||
} | ||
} | ||
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@Override | ||
public String toString() { | ||
return "Node [value=" + value + "]"; | ||
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} | ||
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/** | ||
* 添加结点的方法,递归的形式添加结点,注意需要满足二叉排序树的要求 | ||
* | ||
* @param node | ||
*/ | ||
public void add(Node node) { | ||
if (node == null) { | ||
return; | ||
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} | ||
//判断传入的结点的值,和当前子树的根结点的值关系 | ||
if (node.value < this.value) { | ||
//如果当前结点左子结点为 null | ||
if (this.left == null) { | ||
this.left = node; | ||
} else { | ||
//递归的向左子树添加 | ||
this.left.add(node); | ||
} | ||
} else { //添加的结点的值大于 当前结点的值 | ||
if (this.right == null) { | ||
this.right = node; | ||
} else { | ||
//递归的向右子树添加 | ||
this.right.add(node); | ||
} | ||
} | ||
} | ||
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//中序遍历 | ||
public void infixOrder() { | ||
if (this.left != null) { | ||
this.left.infixOrder(); | ||
} | ||
System.out.println(this); | ||
if (this.right != null) { | ||
this.right.infixOrder(); | ||
} | ||
} | ||
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} | ||
``` | ||
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