java - 在 Java 中实现和可视化二叉树

Question

我必须实现此文件中描绘的二叉树

连同类 Diagramm 和二叉树用于定位。

所以按照文本和图片，我必须为这个二叉树实现一个构造函数，一个获取和插入方法。

public class BinaryTree {

 private Node root = null;

private static class Node {
    private Integer key;
    private String value;
    private Node left = null;
    private Node right = null;

    public Node(Integer key, String value) {
        this.key = key;
        this.value = value;
    }
}

public boolean insert(Integer key, String value) {
    if (root == null) {
        root = new Node(key, value);
        return true;
    } else {
        return insert(root, key, value);
    }
}

private boolean insert(Node node, Integer key, String value) {
    if (key.equals(node.key)) {
        // duplicate
        return false;
    } else if (key < node.key) {
        if (node.left == null) {
            node.left = new Node(key, value);
            return true;
        } else {
            return insert(node.left, key, value);
        }
    } else if (key > node.key) {
        if (node.right == null) {
            node.right = new Node(key, value);
            return true;
        } else {
            return insert(node.right, key, value);
        }
    }
    return false;
}

// not tested, crass assumptions, public domain
public String get(Integer key) {
    return get(root, key); // start search from the root.
}

public String get(Node node, Integer key) {
    String result = null; // Assume key is not found

    if (node.key.equals(key)) { // Key matches? This is the result.
        return node.value;
    } else {
        if (key < node.key && node.left != null) { // key is lower than
                                                    // current node,
                                                    // and there is a left
                                                    // branch, keep
                                                    // search from there.
            result = get(node.left, key);
        } else if (key > node.key && node.right != null) { // key is greater
                                                            // than current
                                                            // node,
                                                            // and there is
                                                            // a left
                                                            // branch,
                                                            // keep search
                                                            // from there.
                                                            // The key >
                                                            // node.key is
                                                            // arguably
                                                            // redundant.
            result = get(node.right, key);
        }
    }

    return result;
}

如何实现正确的测试主要功能？在顶部，我必须借助 graphviz 可视化二叉树并在节点类中添加一个方法，该方法为点代码创建一个字符串。它如何与 Eclipse 配合使用？

Answer 1

get 方法将从树中的特定节点开始，并检查该节点本身是否符合条件。如果是，则返回该值。如果没有，它将推迟到适当的分支并继续搜索。

// not tested, crass assumptions, public domain
public String get(Integer key) {
    return get(root, key); // start search from the root.
}

public String get(Node node, Integer key) {
    String result = null; // Assume key is not found

    if (node.key.equals(key)) { // Key matches? This is the result.
        return node.value;
    } else {
        if (key < node.key && node.left != null) {  // key is lower than current node, 
                                                    // and there is a left branch, keep
                                                    // search from there.
            result = get(node.left, key);
        } else if (key > node.key && node.right != null) {  // key is greater than current node,
                                                            // and there is a left branch,
                                                            // keep search from there.
                                                            // The key > node.key is arguably
                                                            // redundant.
            result = get(node.right, key);
        }
    }

    return result;
}