Question

所有人。

我正在尝试提出一种允许用户在运行时创建树的方法，其分支因子为2，深度不受限制（取决于用户）。

用户必须以 root_node 开头，然后继续到根节点的两个子节点（ left_node 和 right_node ）。之后，左边的孩子成为根，并且用户执行相同的过程并移至右边的孩子。

感谢您提供任何有关实现此目标的帮助。

预先感谢

Answer 1

您是否正在寻找类似的东西？

import java.util.LinkedList;
  import java.util.Queue;

public class BinaryTree {

    Node root;

    public void add(int value) {
        root = addRecursive(root, value);
    }

    private Node addRecursive(Node current, int value) {

        if (current == null) {
            return new Node(value);
        }

        if (value < current.value) {
            current.left = addRecursive(current.left, value);
        } else if (value > current.value) {
            current.right = addRecursive(current.right, value);
        }

        return current;
    }

    public boolean isEmpty() {
        return root == null;
    }

    public int getSize() {
        return getSizeRecursive(root);
    }

    private int getSizeRecursive(Node current) {
        return current == null ? 0 : getSizeRecursive(current.left) + 1 + getSizeRecursive(current.right);
    }

    public boolean containsNode(int value) {
        return containsNodeRecursive(root, value);
    }

    private boolean containsNodeRecursive(Node current, int value) {
        if (current == null) {
            return false;
        }

        if (value == current.value) {
            return true;
        }

        return value < current.value
          ? containsNodeRecursive(current.left, value)
          : containsNodeRecursive(current.right, value);
    }

    public void delete(int value) {
        root = deleteRecursive(root, value);
    }

    private Node deleteRecursive(Node current, int value) {
        if (current == null) {
            return null;
        }

        if (value == current.value) {
            // Case 1: no children
            if (current.left == null && current.right == null) {
                return null;
            }

            // Case 2: only 1 child
            if (current.right == null) {
                return current.left;
            }

            if (current.left == null) {
                return current.right;
            }

            // Case 3: 2 children
            int smallestValue = findSmallestValue(current.right);
            current.value = smallestValue;
            current.right = deleteRecursive(current.right, smallestValue);
            return current;
        }
        if (value < current.value) {
            current.left = deleteRecursive(current.left, value);
            return current;
        }

        current.right = deleteRecursive(current.right, value);
        return current;
    }

    private int findSmallestValue(Node root) {
        return root.left == null ? root.value : findSmallestValue(root.left);
    }

    public void traverseInOrder(Node node) {
        if (node != null) {
            traverseInOrder(node.left);
            System.out.print(" " + node.value);
            traverseInOrder(node.right);
        }
    }

    public void traversePreOrder(Node node) {
        if (node != null) {
            System.out.print(" " + node.value);
            traversePreOrder(node.left);
            traversePreOrder(node.right);
        }
    }

    public void traversePostOrder(Node node) {
        if (node != null) {
            traversePostOrder(node.left);
            traversePostOrder(node.right);
            System.out.print(" " + node.value);
        }
    }

    public void traverseLevelOrder() {
        if (root == null) {
            return;
        }

        Queue<Node> nodes = new LinkedList<>();
        nodes.add(root);

        while (!nodes.isEmpty()) {

            Node node = nodes.remove();

            System.out.print(" " + node.value);

            if (node.left != null) {
                nodes.add(node.left);
            }

            if (node.left != null) {
                nodes.add(node.right);
            }
        }
    }

    class Node {
        int value;
        Node left;
        Node right;

        Node(int value) {
            this.value = value;
            right = null;
            left = null;
        }
    }
}

在运行时创建树

1 个答案: