binary-tree - 在.dot树中强制执行水平节点排序

我正在尝试使用GraphViz为二进制搜索树重新创建示例图。最终应该是这样的:

这是我的第一次尝试:

digraph G {
    nodesep=0.3;
    ranksep=0.2;
    margin=0.1;
    node [shape=circle];
    edge [arrowsize=0.8];
    6 -> 4;
    6 -> 11;
    4 -> 2;
    4 -> 5;
    2 -> 1;
    2 -> 3;
    11 -> 8;
    11 -> 14;
    8 -> 7;
    8 -> 10;
    10 -> 9;
    14 -> 13;
    14 -> 16;
    13 -> 12;
    16 -> 15;
    16 -> 17;
}

但是不幸的是GraphViz不在乎树的水平位置，所以我得到:

如何添加约束，以使顶点的水平位置反射(reflect)其总体顺序？

最佳答案

您可以按照通常的方法添加不可见的节点和不可见的边缘，并按照graphviz FAQ about balanced trees的建议使用边缘权重等。在一些simple cases中，这就足够了。

但是，有一个更好的解决方案:Graphviz附带了一个名为 gvpr (图形模式扫描和处理语言)的工具，该工具可以



而且自Emden R. Gansner did all the work already by creating a script which does layout nicely binary trees以来，这是如何做到的(所有功劳归功于ERG):

将以下gvpr脚本保存到名为tree.gv的文件中:

BEGIN {
  double tw[node_t];    // width of tree rooted at node
  double nw[node_t];    // width of node
  double xoff[node_t];  // x offset of root from left side of its tree
  double sp = 36;       // extra space between left and right subtrees
  double wd, w, w1, w2;
  double x, y, z;
  edge_t e1, e2;
  node_t n;
}
BEG_G {
  $.bb = "";
  $tvtype=TV_postfwd;   // visit root after all children visited
}
N {
  sscanf ($.width, "%f", &w);
  w *= 72;  // convert inches to points
  nw[$] = w;
  if ($.outdegree == 0) {
    tw[$] = w;
    xoff[$] = w/2.0;
  }
  else if ($.outdegree == 1) {
    e1 = fstout($);
    w1 = tw[e1.head];
    tw[$] = w1 + (sp+w)/2.0;
    if (e1.side == "left")
      xoff[$] = tw[$] - w/2.0;
    else
      xoff[$] = w/2.0;
  }
  else {
    e1 = fstout($);
    w1 = tw[e1.head];
    e2 = nxtout(e1);
    w2 = tw[e2.head];
    wd = w1 + w2 + sp;
    if (w > wd)
      wd = w;
    tw[$] = wd;
    xoff[$] = w1 + sp/2.0;
  }
}
BEG_G {
  $tvtype=TV_fwd;   // visit root first, then children
}
N {
  if ($.indegree == 0) {
    sscanf ($.pos, "%f,%f", &x, &y);
    $.pos = sprintf("0,%f", y);
  }
  if ($.outdegree == 0) return;
  sscanf ($.pos, "%f,%f", &x, &y);
  wd = tw[$];
  e1 = fstout($);
  n = e1.head;
  sscanf (n.pos, "%f,%f", &z, &y);
  if ($.outdegree == 1) {
    if (e1.side == "left")
      n.pos = sprintf("%f,%f",  x - tw[n] - sp/2.0 + xoff[n], y);
    else
      n.pos = sprintf("%f,%f", x + sp/2.0 + xoff[n], y);
  }
  else {
    n.pos = sprintf("%f,%f", x - tw[n] - sp/2.0 + xoff[n], y);
    e2 = nxtout(e1);
    n = e2.head;
    sscanf (n.pos, "%f,%f", &z, &y);
    n.pos = sprintf("%f,%f", x + sp/2.0 + xoff[n], y);
  }
}

假设包含该图的点文件称为binarytree.gv，则可以执行以下行:

dot binarytree.gv | gvpr -c -ftree.gv | neato -n -Tpng -o binarytree.png

结果是:

通过在脚本中切换一两行，您甚至可以使单个子节点位于左侧而不是右侧。

关于binary-tree - 在.dot树中强制执行水平节点排序，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/10902745/