c# - N-ary树插入和搜索的复杂度是多少？

Question

我正在用 C# 实现 N-1ry 树。我想知道如何计算以下方法的复杂性。这是我的代码:

结构:

public class Node
{
    public int Value { get; set; }
    public Node Children { get; set; }
    public Node Sibilings { get; set; }

}

这种搜索方法:

public Node search(Node root, int data)
{
    if (root == null)
        return null;

    if (data == root.Value)
        return root;

    Node t = search(root.Children, data);
    if (t == null)
        t = search(root.Sibilings, data);
    return t;
}

这种插​​入方法:

public void Add(int[] data)
{
    Node temp = null;
    temp = search(ROOT, data[0]);

    if (temp == null)
        temp = new Node(data[0]);

    if (this.ROOT == null)
        ROOT = temp;
    Node parent = temp;

    for (int j = 1; j <= this.NoOfChildrens; j++)
    {
        // for first child
        if (j == 1)
        {
            parent.Children = new Node(data[j]);
            parent = parent.Children;
        }
        //for all other childs
        else
        {
            parent.Sibilings = new Node(data[j]);
            parent = parent.Sibilings;
        }

    }

}

程序入口点:

static void Main(string[] args)
{
    NAryTree naryTree = new NAryTree(3);

    // 1st element in each row is node Value,>=2nd....=>value of child

    int[][] data = { new int[] { 1, 2, 3, 4 }, new int[] { 2, 1, 6,0 }, new int[] { 3, 8, 9, 10 }, new int[] { 4, 0, 0, 0 } };

    naryTree.Add(data[0]);
    naryTree.Add(data[1]);
    naryTree.Add(data[2]);
    naryTree.Add(data[3]);
    naryTree.Add(new int[] {10,3,6,4});

    naryTree.preorder(naryTree.ROOT);
    Console.ReadLine();
}

这些方法的 bigO 复杂度是多少？

Answer 1

让我们看看 Search 方法中有什么。它不是二叉树，我们有递归。所以 Search 方法将调用 N 次，直到我们找到所需的值。所以我们可以得出结论，我们有 O(N)，其中 N 是在最后一次迭代中找到值的最大(最差)迭代次数:

public Node search(Node root, int data)
{
    if (root == null)
        return null;

    if (data == root.Value)
        return root;

    Node t = search(root.Children, data);
    if (t == null)
        t = search(root.Sibilings, data);
    return t;
} 

For Addition 方法更简单，因为我们有 for 语句并且没有嵌套循环。所以我们有 O(N) 用于 Addition 方法。

As Wisconsin university says:

So for loops for (i = 0; i < N; i++) { sequence of statements } The loop executes N times, so the sequence of statements also executes N times. Since we assume the statements are O(1), the total time for the for loop is N * O(1), which is O(N) overall.