blocks|key|3093567|text|你可以按照通常的方法添加不可见的节点和不可见的边，并玩边权重等，就像在graphviz+FAQ+about+balanced+trees上建议的那样。在某些simple+cases中，这就足够了。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3093568|但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，它允许|style|BOLD|3093569|3093570|将输入图表复制到其输出，可能会转换其结构和属性、创建新图表或打印任意信息|blockquote|3093571|3093572|3093573|既然是Emden+R.+Gansner+did+all+the+work+already+by+creating+a+script+which+does+layout+nicely+binary+trees，下面是如何做到这一点的(所有功劳都归功于ERG)：|3093574|将以下gvpr脚本保存到名为tree.gv的文件中：|CODE|3093575|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+($.width,+"%25f",+&w);
++w+*=+72;++//+convert+inches+to+points
++nw[$]+=+w;
++if+($.outdegree+==+0)+{
++++tw[$]+=+w;
++++xoff[$]+=+w/2.0;
++}
++else+if+($.outdegree+==+1)+{
++++e1+=+fstout($);
++++w1+=+tw[e1.head];++++
++++tw[$]+=+w1+%2B+(sp%2Bw)/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+%2B+w2+%2B+sp;
++++if+(w+>+wd)
++++++wd+=+w;
++++tw[$]+=+wd;
++++xoff[$]+=+w1+%2B+sp/2.0;
++}
}
BEG_G+{
++$tvtype=TV_fwd;+++//+visit+root+first,+then+children
}
N+{
++if+($.indegree+==+0)+{
++++sscanf+($.pos,+"%25f,%25f",+&x,+&y);
++++$.pos+=+sprintf("0,%25f",+y);
++}
++if+($.outdegree+==+0)+return;
++sscanf+($.pos,+"%25f,%25f",+&x,+&y);
++wd+=+tw[$];
++e1+=+fstout($);
++n+=+e1.head;
++sscanf+(n.pos,+"%25f,%25f",+&z,+&y);
++if+($.outdegree+==+1)+{
++++if+(e1.side+==+"left")
++++++n.pos+=+sprintf("%25f,%25f",++x+-+tw[n]+-+sp/2.0+%2B+xoff[n],+y);
++++else
++++++n.pos+=+sprintf("%25f,%25f",+x+%2B+sp/2.0+%2B+xoff[n],+y);
++}
++else+{
++++n.pos+=+sprintf("%25f,%25f",+x+-+tw[n]+-+sp/2.0+%2B+xoff[n],+y);
++++e2+=+nxtout(e1);
++++n+=+e2.head;
++++sscanf+(n.pos,+"%25f,%25f",+&z,+&y);
++++n.pos+=+sprintf("%25f,%25f",+x+%2B+sp/2.0+%2B+xoff[n],+y);
++}
}|code-block|syntax|javascript|3093576|假设包含图形的点文件名为binarytree.gv，您可以执行以下行：|3093577|dot+binarytree.gv+%7C+gvpr+-c+-ftree.gv+%7C+neato+-n+-Tpng+-o+binarytree.png|3093578|结果是：|3093579|​|3093580|📷|atomic|3093581|3093582|通过在脚本中切换一两行，您甚至可以让单个子节点转到左侧，而不是右侧。|3093583|entityMap|0|LINK|mutability|MUTABLE|url|http://www.graphviz.org/content/FaqBalanceTree|1|https://stackoverflow.com/q/3105923/63733|2|https://mailman.research.att.com/pipermail/graphviz-interest/2010q2/007101.html|3|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/46d77e2d8c8d43004e9de4ec845f89df.png|imageAlt^0|Z|X|0|26|C|1|0|S|4|0|0|0|0|0|3|2S|2|0|E|7|0|0|C|D|0|0|0|0|0|1|3|0|0|0^^$0|@$1|2|3|4|5|6|7|1T|8|@]|9|@$A|1U|B|1V|1|1W]|$A|1X|B|1Y|1|1Z]]|C|$]]|$1|D|3|E|5|6|7|20|8|@$A|21|B|22|F|G]]|9|@]|C|$]]|$1|H|3|-4|5|6|7|23|8|@]|9|@]|C|$]]|$1|I|3|J|5|K|7|24|8|@]|9|@]|C|$]]|$1|L|3|-4|5|6|7|25|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|26|8|@]|9|@]|C|$]]|$1|N|3|O|5|6|7|27|8|@]|9|@$A|28|B|29|1|2A]]|C|$]]|$1|P|3|Q|5|6|7|2B|8|@$A|2C|B|2D|F|R]]|9|@]|C|$]]|$1|S|3|T|5|U|7|2E|8|@]|9|@]|C|$V|W]]|$1|X|3|Y|5|6|7|2F|8|@$A|2G|B|2H|F|R]]|9|@]|C|$]]|$1|Z|3|10|5|U|7|2I|8|@]|9|@]|C|$V|W]]|$1|11|3|12|5|6|7|2J|8|@]|9|@]|C|$]]|$1|13|3|14|5|6|7|2K|8|@]|9|@]|C|$]]|$1|15|3|16|5|17|7|2L|8|@]|9|@$A|2M|B|2N|1|2O]]|C|$]]|$1|18|3|14|5|6|7|2P|8|@]|9|@]|C|$]]|$1|19|3|1A|5|6|7|2Q|8|@]|9|@]|C|$]]|$1|1B|3|-4|5|6|7|2R|8|@]|9|@]|C|$]]]|1C|$1D|$5|1E|1F|1G|C|$1H|1I]]|1J|$5|1E|1F|1G|C|$1H|1K]]|1L|$5|1E|1F|1G|C|$1H|1M]]|1N|$5|1O|1F|1P|C|$1Q|1R|1S|-4]]]]

You could follow the usual approach of adding invisible nodes and invisible edges, and playing with edge weight etc. as proposed on the <a href="http://www.graphviz.org/content/FaqBalanceTree" rel="noreferrer">graphviz FAQ about balanced trees</a>. In some <a href="https://stackoverflow.com/q/3105923/63733">simple cases</a>, this is enough.

But there is a better solution: Graphviz comes with a tool called gvpr (graph pattern scanning and processing language) which allows to

<blockquote>
 copy input graphs to its output, possibly transforming their structure and attributes, creating new graphs, or printing arbitrary information
</blockquote>

And since <a href="https://mailman.research.att.com/pipermail/graphviz-interest/2010q2/007101.html" rel="noreferrer">Emden R. Gansner did all the work already by creating a script which does layout nicely binary trees</a>, heres how to to that (all credit goes to ERG):

Save the following gvpr script into a file called <code>tree.gv</code> :

<pre><code>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", &amp;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($);
 w1 = tw[e1.head]; 
 e2 = nxtout(e1);
 w2 = tw[e2.head]; 
 wd = w1 + w2 + sp;
 if (w &gt; wd)
 wd = w;
 tw[$] = wd;
 xoff[$] = w1 + sp/2.0;
 }
}
BEG_G {
 $tvtype=TV_fwd; // visit root first, then children
}
N {
 if ($.indegree == 0) {
 sscanf ($.pos, "%f,%f", &amp;x, &amp;y);
 $.pos = sprintf("0,%f", y);
 }
 if ($.outdegree == 0) return;
 sscanf ($.pos, "%f,%f", &amp;x, &amp;y);
 wd = tw[$];
 e1 = fstout($);
 n = e1.head;
 sscanf (n.pos, "%f,%f", &amp;z, &amp;y);
 if ($.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", &amp;z, &amp;y);
 n.pos = sprintf("%f,%f", x + sp/2.0 + xoff[n], y);
 }
}
</code></pre>

Assuming your dot file containing the graph is called <code>binarytree.gv</code>, you may execute the following line:

<pre><code>dot binarytree.gv | gvpr -c -ftree.gv | neato -n -Tpng -o binarytree.png
</code></pre>

The result is:

<img src="https://i.stack.imgur.com/st6rl.png" alt="Binary tree nicely layouted with graphiv and a gvpr script thanks to Emden R. Gansner">

By switching around a line or two in the script, you'll even get to have the single child nodes go to the left instead of the right side.

I am trying to recreate an example diagram for a binary search tree with GraphViz. This is how it should look in the end:

<img src="https://i.stack.imgur.com/xiFxO.png" alt="enter image description here">

This is my first attempt:

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

But unfortunately GraphViz doesn't care about the horizontal positions of the tree, so I get:

<img src="https://i.stack.imgur.com/QQ43l.png" alt="enter image description here">

How can I add constraints so that the horizontal positions of the vertices reflects their total ordering?

Enforcing horizontal node ordering in a .dot tree

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试用GraphViz重新创建一个二叉树的示例图。下面是它最终应该是什么样子：​​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]...

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