Question

所以我实际上有两个问题。首先，我可以指定顶点显示的顺序吗？我的意思是，当我试图想象顶点的直线时，有时它会先到最后，有时会持续到第一个。这是我的代码片段。还有什么更好的方法来制作直线而不是将垂直尺寸改为低点？

Tree<GraphNode, String> g1 = q.transformToGraph();

Layout<GraphNode, String> layout = new ISOMLayout<GraphNode, String>(g1);
    layout.setSize(new Dimension(1600, 25));
VisualizationViewer<GraphNode, String> vv = new VisualizationViewer<GraphNode, String>(layout);
vv.getRenderContext().setEdgeShapeTransformer(new EdgeShape.Line());
VertexLabelAsShapeRenderer<GraphNode, String> vlasr = new VertexLabelAsShapeRenderer<GraphNode, String>(vv.getRenderContext());
vv.getRenderContext().setVertexShapeTransformer(vlasr);
vv.getRenderContext().setVertexLabelTransformer(
    new ChainedTransformer<GraphNode, String>(
            new Transformer[]{new MyLabeller()}));
vv.getRenderer().setVertexLabelRenderer(vlasr);
return vv;

transformToGraph函数返回DirectedSparseGraph。

还有什么办法可以不编写我自己的布局来制作漂亮的二叉树可视化。特别是BST，因为TreeLayout使边缘长度不同并且交换了一些边缘的边（左边是右边而另一边是）

Tree<GraphNode, String> g1 = (DelegateTree) q.transformToGraph();

    Layout<GraphNode, String> layout = new DynamicTreeLayout<GraphNode, String>(g1, 100, 100);
layout.setSize(new Dimension(1600, 100));       VisualizationViewer<GraphNode, String> vv = new VisualizationViewer<GraphNode, String>(layout);
vv.getRenderContext().setEdgeShapeTransformer(new EdgeShape.Line());
VertexLabelAsShapeRenderer<GraphNode, String> vlasr = new VertexLabelAsShapeRenderer<GraphNode, String>(vv.getRenderContext());
vv.getRenderContext().setVertexShapeTransformer(vlasr);
vv.getRenderContext().setVertexLabelTransformer(
    new ChainedTransformer<GraphNode, String>(
        new Transformer[]{new MyLabeller()}));
vv.getRenderer().setVertexLabelRenderer(vlasr);
return vv;

这里transformToGraph将我的BST更改为DelegateTree。

以下是

@Override
    public DelegateTree<GraphNode, String> transformToGraph() {
        DelegateTree<GraphNode, String> graph = new DelegateTree<GraphNode, String>();
        if (root == null) {
            graph.setRoot(null);
            return graph;
        }
        GraphNode rootNode = new GraphNode(root.getValue());
        graph.setRoot(rootNode);
        if (root.getLeft() == root.getRight()) {
            return graph;
        }
        addNodeToGraph(rootNode, root.getLeft(), graph);
        addNodeToGraph(rootNode, root.getRight(), graph);
        return graph;
    }
    protected void addNodeToGraph(GraphNode parent, TreeNode node, DelegateTree<GraphNode, String> graph) {
        if (node == null) {
            return;
        }
        GraphNode curr = new GraphNode(node.getValue());
        graph.addChild(parent.getValue() + node.getValue(), parent, curr);
        addNodeToGraph(curr, node.getLeft(), graph);
        addNodeToGraph(curr, node.getRight(), graph);
    }

Answer 1

图中的节点或节点的邻居/后继者/前任者的迭代次序由图实现确定。有些实现强制执行一致的排序，有些则不执行。一个实现：http://jung.sourceforge.net/doc/api/edu/uci/ics/jung/graph/DirectedOrderedSparseMultigraph.html

如果你想要一个折线图，你很可能会想要编写自己的Layout实现;不过应该很容易。（一般图形的良好布局算法要困难得多，但对于一个简单的链条，它应该基本上是微不足道的。）

JUNG顶点，JUNG二叉树的静态顺序

1 个答案: