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134 lines
No EOL
3.9 KiB
C#
134 lines
No EOL
3.9 KiB
C#
/**
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* File: array_binary_tree.cs
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* Created Time: 2023-07-20
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* Author: hpstory (hpstory1024@163.com)
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*/
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namespace hello_algo.chapter_tree;
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/* 数组表示下的二叉树类 */
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public class ArrayBinaryTree {
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private List<int?> tree;
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/* 构造方法 */
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public ArrayBinaryTree(List<int?> arr) {
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tree = new List<int?>(arr);
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}
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/* 节点数量 */
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public int size() {
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return tree.Count;
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}
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/* 获取索引为 i 节点的值 */
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public int? val(int i) {
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// 若索引越界,则返回 null ,代表空位
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if (i < 0 || i >= size())
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return null;
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return tree[i];
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}
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/* 获取索引为 i 节点的左子节点的索引 */
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public int left(int i) {
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return 2 * i + 1;
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}
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/* 获取索引为 i 节点的右子节点的索引 */
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public int right(int i) {
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return 2 * i + 2;
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}
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/* 获取索引为 i 节点的父节点的索引 */
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public int parent(int i) {
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return (i - 1) / 2;
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}
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/* 层序遍历 */
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public List<int> levelOrder() {
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List<int> res = new List<int>();
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// 直接遍历数组
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for (int i = 0; i < size(); i++) {
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if (val(i).HasValue)
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res.Add(val(i).Value);
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}
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return res;
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}
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/* 深度优先遍历 */
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private void dfs(int i, string order, List<int> res) {
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// 若为空位,则返回
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if (!val(i).HasValue)
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return;
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// 前序遍历
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if (order == "pre")
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res.Add(val(i).Value);
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dfs(left(i), order, res);
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// 中序遍历
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if (order == "in")
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res.Add(val(i).Value);
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dfs(right(i), order, res);
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// 后序遍历
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if (order == "post")
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res.Add(val(i).Value);
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}
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/* 前序遍历 */
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public List<int> preOrder() {
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List<int> res = new List<int>();
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dfs(0, "pre", res);
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return res;
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}
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/* 中序遍历 */
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public List<int> inOrder() {
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List<int> res = new List<int>();
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dfs(0, "in", res);
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return res;
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}
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/* 后序遍历 */
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public List<int> postOrder() {
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List<int> res = new List<int>();
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dfs(0, "post", res);
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return res;
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}
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}
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public class array_binary_tree {
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[Test]
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public void Test() {
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// 初始化二叉树
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// 这里借助了一个从数组直接生成二叉树的函数
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List<int?> arr = new List<int?> { 1, 2, 3, 4, null, 6, 7, 8, 9, null, null, 12, null, null, 15 };
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TreeNode root = TreeNode.ListToTree(arr);
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Console.WriteLine("\n初始化二叉树\n");
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Console.WriteLine("二叉树的数组表示:");
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Console.WriteLine(arr.PrintList());
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Console.WriteLine("二叉树的链表表示:");
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PrintUtil.PrintTree(root);
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// 数组表示下的二叉树类
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ArrayBinaryTree abt = new ArrayBinaryTree(arr);
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// 访问节点
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int i = 1;
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int l = abt.left(i);
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int r = abt.right(i);
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int p = abt.parent(i);
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Console.WriteLine("\n当前节点的索引为 " + i + " ,值为 " + abt.val(i));
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Console.WriteLine("其左子节点的索引为 " + l + " ,值为 " + (abt.val(l).HasValue ? abt.val(l) : "null"));
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Console.WriteLine("其右子节点的索引为 " + r + " ,值为 " + (abt.val(r).HasValue ? abt.val(r) : "null"));
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Console.WriteLine("其父节点的索引为 " + p + " ,值为 " + (abt.val(p).HasValue ? abt.val(p) : "null"));
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// 遍历树
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List<int> res = abt.levelOrder();
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Console.WriteLine("\n层序遍历为:" + res.PrintList());
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res = abt.preOrder();
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Console.WriteLine("前序遍历为:" + res.PrintList());
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res = abt.inOrder();
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Console.WriteLine("中序遍历为:" + res.PrintList());
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res = abt.postOrder();
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Console.WriteLine("后序遍历为:" + res.PrintList());
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}
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} |