C#数据结构-线索化二叉树
为什么线索化二叉树?
对于二叉树的遍历,我们知道每个节点的前驱与后继,但是这是建立在遍历的基础上,否则我们只知道后续的左右子树。现在我们充分利用二叉树左右子树的空节点,分别指向当前节点的前驱、后继,便于快速查找树的前驱后继。
不多说,直接上代码:
/// <summary>
/// 线索二叉树 节点
/// </summary>
/// <typeparam name="T"></typeparam>
public class ClueTreeNode<T>
{
/// <summary>
/// 内容
/// </summary>
public T data { get; set; }
/// <summary>
/// 左树
/// </summary>
public ClueTreeNode<T> leftNode { get; set; }
/// <summary>
/// 右树
/// </summary>
public ClueTreeNode<T> rightNode { get; set; }
/// <summary>
/// 0 标识左树 1 标识 当前节点的前驱
/// </summary>
public int leftTag { get; set; }
/// <summary>
/// 0标识右树 1 标识 当前节点的后继
/// </summary>
public int rightTag { get; set; }
public ClueTreeNode()
{
data = default(T);
leftNode = null;
rightNode = null;
}
public ClueTreeNode(T item)
{
data = item;
leftNode = null;
rightNode = null;
}
}
/// <summary>
/// 线索化 二叉树
///
/// 为什么线索化二叉树?
/// 第一:对于二叉树,如果有n个节点,每个节点有指向左右孩子的两个指针域,所以一共有2n个指针域。
/// 而n个节点的二叉树一共有n-1条分支线数,也就是说,其实是有 2n-(n-1) = n+1个空指针。
/// 这些空间不存储任何事物,白白浪费内存的资源。
/// 第二:对于二叉树的遍历,我们知道每个节点的前驱与后继,但是这是建立在遍历的基础上。
/// 否则我们只知道后续的左右子树。
/// 第三:对于二叉树来说,从结构上来说是单向链表,引入前驱后继后,线索化二叉树可以认为是双向链表。
/// </summary>
/// <typeparam name="T"></typeparam>
public class ClueBinaryTree<T>
{
/// <summary>
/// 树根节
/// </summary>
private ClueTreeNode<T> head { get; set; }
/// <summary>
/// 线索化时作为前驱转存
/// </summary>
private ClueTreeNode<T> preNode { get; set; }
public ClueBinaryTree(){
head = new ClueTreeNode<T>();
}
public ClueBinaryTree(T val){
head = new ClueTreeNode<T>(val);
}
public ClueTreeNode<T> GetRoot(){
return head;
}
/// <summary>
/// 插入左节点
/// </summary>
/// <param name="val"></param>
/// <param name="node"></param>
/// <returns></returns>
public ClueTreeNode<T> AddLeftNode(T val, ClueTreeNode<T> node){
if (node == null)
throw new ArgumentNullException("参数错误");
ClueTreeNode<T> treeNode = new ClueTreeNode<T>(val);
ClueTreeNode<T> childNode = node.leftNode;
treeNode.leftNode = childNode;
node.leftNode = treeNode;
return treeNode;
}
/// <summary>
/// 插入右节点
/// </summary>
/// <param name="val"></param>
/// <param name="node"></param>
/// <returns></returns>
public ClueTreeNode<T> AddRightNode(T val, ClueTreeNode<T> node){
if (node == null)
throw new ArgumentNullException("参数错误");
ClueTreeNode<T> treeNode = new ClueTreeNode<T>(val);
ClueTreeNode<T> childNode = node.rightNode;
treeNode.rightNode = childNode;
node.rightNode = treeNode;
return treeNode;
}
/// <summary>
/// 删除当前节点的 左节点
/// </summary>
/// <param name="node"></param>
/// <returns></returns>
public ClueTreeNode<T> DeleteLeftNode(ClueTreeNode<T> node){
if (node == null || node.leftNode == null)
throw new ArgumentNullException("参数错误");
ClueTreeNode<T> leftChild = node.leftNode;
node.leftNode = null;
return leftChild;
}
/// <summary>
/// 删除当前节点的 右节点
/// </summary>
/// <param name="node"></param>
/// <returns></returns>
public ClueTreeNode<T> DeleteRightNode(ClueTreeNode<T> node){
if (node == null || node.rightNode == null)
throw new ArgumentNullException("参数错误");
ClueTreeNode<T> rightChild = node.rightNode;
node.rightNode = null;
return rightChild;
}
/// <summary>
/// 中序遍历线索化二叉树
/// </summary>
public void MiddlePrefaceTraversal(){
ClueTreeNode<T> node = head;
while (node != null)
{
//判断是否是
while (node.leftTag == 0)
{
node = node.leftNode;
}
Console.Write($" {node.data}");
while (node.rightTag == 1)
{
node = node.rightNode;
Console.Write($" {node.data}");
}
node = node.rightNode;
}
}
/// <summary>
/// 线索化二叉树
/// </summary>
/// <param name="node"></param>
public void MiddleClueNodes(ClueTreeNode<T> node){
if (node == null)
{
return;
}
//线索化左子树
MiddleClueNodes(node.leftNode);
//当左树为空时,指向前驱,标识为 1
if (node.leftNode == null)
{
node.leftNode = preNode;
node.leftTag = 1;
}
//如果 前驱的右树不为空
if (preNode != null && preNode.rightNode == null)
{
preNode.rightNode = node;
preNode.rightTag = 1;
}
preNode = node;
//线索化右子树
MiddleClueNodes(node.rightNode);
}
}
线索化二叉树的过程(中序遍历):
现在我们测试:
//创建树
ClueBinaryTree<string> clueBinaryTree = new ClueBinaryTree<string>("A");
ClueTreeNode<string> tree1 = clueBinaryTree.AddLeftNode("B", clueBinaryTree.GetRoot());
ClueTreeNode<string> tree2 = clueBinaryTree.AddRightNode("C", clueBinaryTree.GetRoot());
ClueTreeNode<string> tree3 = clueBinaryTree.AddLeftNode("D", tree1);
clueBinaryTree.AddRightNode("E", tree1);
clueBinaryTree.AddLeftNode("F", tree2);
clueBinaryTree.AddRightNode("G", tree2);
clueBinaryTree.MiddleClueNodes(clueBinaryTree.GetRoot());
Console.Write("中序遍历");
clueBinaryTree.MiddlePrefaceTraversal();
打印结果:
中序遍历 D B E A F C G