使用Graphviz画一颗二叉树

概述

发现有一个工具叫Graphviz，类似于markdown，它使用dot标记语言来编写，能将其转换为图形。

准备工作

（非必需，文末有精简版下载地址）
下载并安装Graphviz
将安装目录/bin添加到环境变量

步骤

编写脚本

digraph binTree{
    node[shape=circle,fontsize=30,color=blue];
    root->a[side=left];
    root->b[side=right];
    a->c[side=left];
    a->d[side=right];
    b->e[side=left];
    e->f[side=right];
}}

直接使用dot -Tpng -o tree.png tree.dot转换，结果为：

左右节点不明显。

使用优化模板转换

binarytree.gvpr内容如下：

// from Emden Gansner
// https://mailman.research.att.com/pipermail/graphviz-interest/2010q2/007101.html
// requires GraphViz 2.28.0 (fails with 2.26.3 at least)
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);
  }
}