C#数据结构-线索化二叉树

Posted 2021-04-09 DotNet且行且珍惜

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