Hcfyk

I am trying to draw a tree using tikzpicture like this:

documentclassarticle
usepackagetikz
usetikzlibrarymath
begindocument
begintikzpicture
 tikzstylenode=[circle, fill=blue!25, minimum size=0.1 cm];
 foreach ilayer in 0,...,3 
 tikzmath nnodes = 3 ^ ilayer; 
 tikzmath leftnum = 1 - floor(nnodes / 2) - 1; 
 tikzmath rightnum = nnodes - floor(nnodes / 2) - 1; 
 foreach isibling in leftnum,...,rightnum 
 tikzmath d = 3 ^ (- ilayer) * 15; 
 tikzmath x = isibling * d; 
 tikzmath y = - ilayer * 2; 
 node[node] (node_ilayer_isibling) at (x cm, y cm) isibling;
 
 
endtikzpicture
enddocument

I get the tree like the following picture. The texts are nodes' isibling within each layer. Most nodes are integers, but all leftmost nodes are not.

You could just tell TikZ explicitly that you want an integer.

documentclass[tikz,border=3.14mm]standalone
usetikzlibrarymath
begindocument
begintikzpicture
 tikzsetmynode/.style=circle, fill=blue!25, minimum size=0.1 cm
 foreach ilayer in 0,...,3 
 tikzmath nnodes = 3 ^ ilayer; 
 tikzmath leftnum = int(1 - floor(nnodes / 2) - 1); 
 tikzmath rightnum = nnodes - floor(nnodes / 2) - 1; 
 foreach isibling in leftnum,...,rightnum 
 tikzmath d = 3 ^ (- ilayer) * 15; 
 tikzmath x = isibling * d; 
 tikzmath y = - ilayer * 2; 
 node[mynode] (node_ilayer_isibling) at (x cm, y cm) isibling;
 
 
endtikzpicture
enddocument

Or

documentclass[tikz,border=3.14mm]standalone
usetikzlibrarymath
begindocument
begintikzpicture
 tikzsetmynode/.style=circle, fill=blue!25, minimum size=0.1 cm
 foreach ilayer in 0,...,3 
 tikzmath int nnodes,leftnum,rightnum;
 nnodes = 3 ^ ilayer;
 leftnum = 1 - floor(nnodes / 2) - 1; 
 rightnum = nnodes - floor(nnodes / 2) - 1; 
 foreach isibling in leftnum,...,rightnum 
 tikzmath d = 3 ^ (- ilayer) * 15; 
 x = isibling * d; 
 y = - ilayer * 2; 
 node[mynode] (node_ilayer_isibling) at (x cm, y cm) isibling;
 
 
endtikzpicture
enddocument

In principle you do not need the math library here.

documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture
 tikzsetmynode/.style=circle, fill=blue!25, minimum size=0.1 cm
 foreach ilayer [evaluate=ilayer as nnodes using int(3 ^ ilayer),
 evaluate=nnodes as leftnum using int(1 - floor(nnodes / 2) - 1),
 evaluate=nnodes as rightnum using int(nnodes - floor(nnodes / 2) - 1)]
 in 0,...,3 
 foreach isibling 
 [evaluate=ilayer as d using 3 ^ (- ilayer) * 15,
 evaluate=isibling as x using isibling * d,
 evaluate=ilayer as y using - ilayer * 2]
 in leftnum,...,rightnum 
 node[mynode] (node_ilayer_isibling) at (x cm, y cm) 
 isibling;
 
 
endtikzpicture
enddocument

As @marmot said you do not need tikzmath here, but if you use it you can do it in more efficient way :

• You can have a single tikzmath command with loops inside it.


• You can declare your integer variables as int so you do not need to do int() afterward.


• As nnodes is odd you do not need separate rightnum and leftnum as rightnum = - leftnum;


• Why you use 1-floor(nnodes/2)-1 in place of -floor(nnodes/2) ?


• The value d can be calculated in the outer loop.


• Instead of using x=isibling*d you can say [x=d cm] and then use isibling asx. And in the same way y can be replaced by ilayer using [y=-2cm].

So here is my proposal :

documentclass[tikz,border=7pt]standalone
usetikzlibrarymath
begindocument
begintikzpicture
 tikzstylenode=[circle, fill=blue!25, minimum size=0.1 cm];
 tikzmath
 int ilayer,nnodes,rightnum,isibling;
 nnodes = 1;
 for ilayer in 0,...,3
 rightnum = (nnodes-1)/2;
 d = 15/nnodes;
 for isibling in -rightnum,...,rightnum
 
 path[x=d cm,y=-2cm]
 node[node] (node_ilayer_isibling) at (isibling, ilayer) isibling;
 ;
 ;
 nnodes = 3*nnodes;
 ;
 
endtikzpicture
enddocument