Question

我尝试为静态二进制树创建一个数据结构，其中包含一个静态根值和两个子节点。我试图让它为任意数量的子值动态化。如何使用静态根节点执行此操作。如果我采用myArray = {3,11,8,18,21,36,1}，我该如何实现。任何没有复杂代码更改的简单代码都会有所帮助。

class Program
{
    static void Main(string[] args)
    {
        TreeNode rootNode = new TreeNode();
        rootNode.value = 9;

        int[] myArray = { 3, 11 };

        for (int i = 0; i < myArray.Length; i++)
        {
            if (myArray[i] > rootNode.value)
            {
                //add to right node
                TreeNode right = new TreeNode();
                right.value = myArray[i];
                rootNode.rightNode = right;

            }
            else
            {
                //add to left node
                TreeNode left = new TreeNode();
                left.value = myArray[i];
                rootNode.leftNode = left;
            }
        }
    }
}

class TreeNode
{
    public int value { get; set; }
    public TreeNode leftNode { get; set; }
    public TreeNode rightNode { get; set; }

}

Answer 1

嘿，这段代码运行正常但你可以根据需要进一步改进。对不起，但我不得不做一点冗长。希望你理解代码。这并不困难。顺便提一句，代码在java中。只需将语法更改为c＃。

class Program
{
    static void Main(string[] args)
    {
        Scanner a = new Scanner(System.in);
        System.out.println("Enter the number of childs!");
        int input = a.nextInt();
        TreeNode rootNode = new TreeNode();
        TreeNode parent = rootNode;
        rootNode.value = 9;

        parent.childNodes = new TreeNode[input];
        for(int i = 0; i< input; i++){
            parent.childNodes[i] = new TreeNode();
            parent.childNodes[i].value = 0;
        }
        parent.hasChild = true;
        int count = 1;
        int startingIndex = 0;
        int EndingIndex = input - 1;
        int next = 0;

        int[] myArray = { 19, 11, 12, 13 ,14, 15 };

        for (int i = 0; i < myArray.length; i++)
        {
            if(count <= input){
                if (myArray[i] > parent.value)
                {
                    //add to right node
                    parent.childNodes[EndingIndex].value = myArray[i];
                    EndingIndex--;
                    count++;
                }

                else
                {
                    //add to the left node
                    parent.childNodes[startingIndex].value = myArray[i];

                    startingIndex++;
                    count++;
                }
            }
            else{
                parent = parent.childNodes[next];
                parent.childNodes = new TreeNode[input];
                for(int j = 0; j< input; j++){
                    parent.childNodes[j] = new TreeNode();
                    parent.childNodes[j].value = 0;
                }
                parent.hasChild = true;
                next++;
                count = 1;
                i--;
                startingIndex = 0;
                EndingIndex = input - 1;
                next = 0;
            }

        }

        parent = rootNode;
        TreeNode grandparent = parent;
        System.out.println("root Node: " + parent.value);
        next = 0;
        int childs = 1;
        while(parent.hasChild == true){
            for(int i=0; i<input; i++){

                if(parent.childNodes[i].value != 0){
                    System.out.print("child " + childs + " : ");
                    childs++;
                    System.out.print(parent.childNodes[i].value);
                    System.out.println();
                }
            }
            childs = 1;
            System.out.println();
            parent = grandparent.childNodes[next];
            next++;
        }
    }
}

class TreeNode
{
    public int value;
    TreeNode[] childNodes;
    boolean hasChild = false;

}

具有静态根节点

1 个答案: