C# AVL树(平衡二叉树)的实现
Posted 不如詩啊
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了C# AVL树(平衡二叉树)的实现相关的知识，希望对你有一定的参考价值。
C# AVL树(平衡二叉树)的实现
- 平衡二叉树的节点类
/// <summary>
/// 平衡二叉树的节点
/// </summary>
public class AVLNode
public int data; //数据
public int bF; //平衡因子
public AVLNode lC; //左孩子
public AVLNode rC; //右孩子
public AVLNode parent; //父节点
public AVLNode(int data)
this.data = data;
- 平衡二叉树类
/// <summary>
/// 平衡二叉树
/// </summary>
class AVLTree
private AVLNode root; //根节点
private int count;
public int Count get => count; private set => count = value;
/// <summary>
/// AVL树的插入方法
/// </summary>
/// <param name="data"></param>
public void Insert(int data)
#region 找到插入点
if (root == null) //若根节点为空, 则将新节点插入到根节点处
root = new AVLNode(data);
return;
AVLNode parent = root; //表示当前节点的父节点
AVLNode curNode = root; //表示当前节点
while (curNode != null)
parent = curNode;
if (data < parent.data) //若data小于当前节点, 则当前节点为当前节点的左孩子
curNode = parent.lC;
else
curNode = parent.rC;
//此时已找到空节点
curNode = new AVLNode(data);
curNode.parent = parent;
if (data < parent.data)
parent.lC = curNode;
else
parent.rC = curNode;
#endregion
//每插入一个节点便要计算平衡因子
//平衡因子为左子树高度减去右子树高度
while (parent != null)
if (parent.lC == curNode) //若新节点在parent的左节点上, 则平衡因子+1
parent.bF++;
else //若新节点在parent的右节点上, 则平衡因子-1
parent.bF--;
//接下来根据平衡因子判断当前父节点是否平衡
if (parent.bF == 0)//若parent的平衡因子为0, 则表示当前为平衡状态, 不需要变更
break;
else if (parent.bF == -1 || parent.bF == 1) //若parent的平衡因子为1或-1, 则需要回溯, 验证上一个节点是否平衡
curNode = parent;
parent = curNode.parent;
else //若parent的平衡因子为2或-2, 则需要重新平衡二叉树
if (parent.bF == 2) //若左子树高于右子树
if (curNode.bF == 1) //LL形
RotateLL(parent); //平衡操作
else //LR形
RotateLR(parent); //平衡操作
else
if (curNode.bF == -1) //RR形
RotateRR(parent); //平衡操作
else //RL形
RotateRL(parent); //平衡操作
break;
count++;
public void RotateLL(AVLNode parent)
/* G G
* 丨 丨
* P Cur
* / \\ / \\
* Cur X1 C P
* / \\ / \\ / \\
* C X2 ===> x3 x4 X2 x1
* / \\
* X3 X4
*/
AVLNode grandParent = parent.parent;
AVLNode curNode = parent.lC;
if (grandParent != null) //若存在G, 则将cur改为g的孩子
if (grandParent.lC == parent)
grandParent.lC = curNode;
curNode.parent = grandParent;
grandParent.bF--;
else
grandParent.rC = curNode;
curNode.parent = grandParent;
grandParent.bF++;
else //若不存在G
curNode.parent = null;
root = curNode;
parent.lC = curNode.rC; //将cur的右孩子X2改为P的左孩子
if (parent.lC != null)
parent.lC.parent = parent; //将P改为X2的parent
curNode.rC = parent; //将P改为cur的右孩子
parent.parent = curNode; //将cur改为P的parent
curNode.bF--;
parent.bF--;
public void RotateLR(AVLNode parent)
/* G G
* 丨 丨
* P C
* / \\ / \\
* Cur X1 Cur P
* / \\ / \\ / \\
* x2 C ===> x2 x3 X4 x1
* / \\
* X3 X4
*/
AVLNode grandParent = parent.parent;
AVLNode curNode = parent.lC;
AVLNode c = curNode.rC;
if (grandParent != null) //若存在G, 则将c改为g的孩子
if (grandParent.lC == parent)
grandParent.lC = c;
c.parent = grandParent;
grandParent.bF--;
else
grandParent.rC = c;
c.parent = grandParent;
grandParent.bF++;
else //若不存在G
c.parent = null;
root = c;
//将cur的右孩子改为x3
curNode.rC = c.lC;
if (curNode.rC != null)
curNode.rC.parent = curNode;
//将p的左孩子改为X4
parent.lC = c.rC;
if (parent.lC != null)
parent.lC.parent = parent;
//将c的左孩子改为cur
c.lC = curNode;
curNode.parent = c;
//将c的右孩子改为P
c.rC = parent;
parent.parent = c;
parent.bF = 0;
curNode.bF++;
c.bF = curNode.bF;
public void RotateRR(AVLNode parent)
/* G G
* 丨 丨
* P Cur
* / \\ / \\
* X1 Cur P C
* / \\ / \\ / \\
* X2 C ===> x1 x2 X3 x4
* / \\
* X3 X4
*/
AVLNode grandParent = parent.parent;
AVLNode curNode = parent.rC;
if (grandParent != null) //若存在G, 则将cur改为g的孩子
if (grandParent.lC == parent)
grandParent.lC = curNode;
curNode.parent = grandParent;
grandParent.bF--;
else
grandParent.rC = curNode;
curNode.parent = grandParent;
grandParent.bF++;
else //若不存在G
curNode.parent = null;
root = curNode;
parent.rC = curNode.lC; //将cur的左孩子X2改为P的右孩子
if (parent.rC != null)
parent.rC.parent = parent; //将P改为X2的parent
curNode.lC = parent; //将P改为cur的左孩子
parent.parent = curNode; //将cur改为P的parent
curNode.bF++;
parent.bF++;
public void RotateRL(AVLNode parent)
/* G G
* 丨 丨
* P C
* / \\ / \\
* X1 Cur P Cur
* / \\ / \\ / \\
* C x2 ===> x1 x3 X4 x2
* / \\
* X3 X4
*/
AVLNode grandParent = parent.parent;
AVLNode curNode = parent.rC;
AVLNode c = curNode.lC;
if (grandParent != null) //若存在G, 则将c改为g的孩子
if (grandParent.lC == parent)
grandParent.lC = c;
c.parent = grandParent;
grandParent.bF--;
else
grandParent.rC = c;
c.parent = grandParent;
grandParent.bF++;
else //若不存在G
c.parent = null;
root = c;
//将cur的左孩子改为x4
curNode.lC = c.rC;
if (curNode.lC != null)
curNode.lC.parent = curNode;
//将p的右孩子改为X3
parent.rC = c.lC;
if (parent.rC != null)
parent.rC.parent = parent;
//将c的右孩子改为cur
c.rC = curNode;
curNode.parent = c;
//将c的左孩子改为P
c.lC = parent;
parent.parent = c;
parent.bF = 0;
curNode.bF--;
c.bF = -curNode.bF;
/// <summary>
/// 中序遍历
/// </summary>
public void InOrder()
_InOrder(root);
/// <summary>
/// 中序遍历(递归)
/// </summary>
/// <param name="node"></param>
private void _InOrder(AVLNode node)
if (node == null)
return;
_InOrder(node.lC);
Console.Write(node.data + " ");
_InOrder(node.rC);
- 主函数
class Program
static void Main(string[] args)
AVLTree aVLTree = new AVLTree();
aVLTree.Insert(16);
aVLTree.Insert(3);
aVLTree.Insert(7);
aVLTree.Insert(11);
aVLTree.Insert(9);
aVLTree.Insert(26);
aVLTree.Insert(18);
aVLTree.Insert(14);
aVLTree.Insert(15);
aVLTree.InOrder();
Console.ReadKey();
- 运行结果
3 7 9 11 14 15 16 18 26
以上是关于C# AVL树(平衡二叉树)的实现的主要内容，如果未能解决你的问题，请参考以下文章