How can I achieve the drawing of a depth-first traversal of a binary tree like in this picture ?
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I think you might have a look to: Highlighting some nodes of a TikZ binomial tree and Hobby path realization in convex hull approach – Claudio Fiandrino Sep 30 '13 at 17:03
2 Answers
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\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc, arrows}
\begin{document}
\begin{tikzpicture}[->,>=stealth',every node/.style={circle,draw},level 1/.style={sibling distance=50mm},level 2/.style={sibling distance=20mm},level 3/.style={sibling distance=12mm},
%scale=0.7, transform shape
]
\node (nA){A}
child { node (nB) {B}
child { node (nD) {D}
child { node (nH) {H} }
}
child { node (nE) {E}
child { node (nI) {I} }
child { node (nJ) {J} }
}
}
child { node (nC) {C}
child { node (nF) {F}
child { node (nK) {K} }
child { node (nL) {L} }
child { node (nM) {M} }
}
child { node (nG) {G} }
};
\draw[->,blue,rounded corners,dashed,line width=0.7pt]
($(nA) + (-0.4,0.2)$) --
($(nB) +(-0.3,0.4)$) --
($(nB) +(-0.6,0.0)$) --
($(nD) +(-0.4,0.3)$) --
($(nD) +(-0.5,0.0)$) --
($(nH) +(-0.5,0.0)$) --
($(nH) +(-0.4,-0.35)$) --
($(nH) +(0.0,-0.5)$) --
($(nH) +(0.4,-0.35)$) --
($(nH) +(0.5,0.0)$) --
% ($(nD) +(0.45,-0.2)$) --
($(nD) +(0.45,0.0)$) --
($(nB) +(0.0,-0.4)$) --
($(nE) +(-0.45,0.0)$) --
($(nI) +(-0.45,0.0)$) --
($(nI) +(-0.35,-0.35)$) --
($(nI) +(0.0,-0.45)$) --
($(nI) +(0.35,-0.35)$) --
($(nI) +(0.4,0.0)$) --
($(nE) +(0.0,-0.4)$) --
($(nJ) +(-0.45,0.0)$) --
($(nJ) +(-0.35,-0.35)$) --
($(nJ) +(0.0,-0.45)$) --
($(nJ) +(0.35,-0.35)$) --
($(nJ) +(0.45,0.0)$) --
($(nE) +(0.4,0.2)$) --
($(nB) +(0.4,0.0)$) --
($(nA) +(0.0,-0.4)$) --
($(nC) +(-0.4,0.0)$) --
% ($(nF) +(-0.6,0.0)$) --
($(nK) +(-0.5,0.1)$) --
($(nK) +(-0.4,-0.35)$) --
($(nK) +(0.0,-0.5)$) --
($(nK) +(0.4,-0.3)$) --
($(nF) +(-0.15,-0.4)$) --
($(nL) +(-0.5,0.0)$) --
($(nL) +(-0.4,-0.35)$) --
($(nL) +(0.0,-0.5)$) --
($(nL) +(0.4,-0.35)$) --
($(nL) +(0.5,0.0)$) --
($(nF) +(0.15,-0.4)$) --
($(nM) +(-0.5,0.0)$) --
($(nM) +(-0.4,-0.35)$) --
($(nM) +(0.0,-0.5)$) --
($(nM) +(0.4,-0.35)$) --
($(nM) +(0.5,0.2)$) --
($(nF) +(0.4,0.0)$) --
($(nC) +(0.0,-0.4)$) --
($(nG) +(-0.5,0.0)$) --
($(nG) +(-0.4,-0.35)$) --
($(nG) +(0.0,-0.5)$) --
($(nG) +(0.4,-0.35)$) --
($(nG) +(0.5,0.1)$) --
($(nC) +(0.6,0.0)$) --
($(nC) +(0.3,0.4)$) --
($(nA) + (0.4,0.2)$);
\end{tikzpicture}
\end{document}
Jake
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Humberto Longo
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1Very nice! Could you maybe edit your answer to include some information on how you generated the path (is it "hand drawn")? – Jake Oct 14 '13 at 13:40
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3
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3
This solution is not as pretty as the hand-drawn version. However, it is somewhat automatic. That said, it will certainly produce unwanted results if a tree differs too significantly from the one I tested it on.
The solution uses forest (of course). I wanted to get intersections using the intersections library in order to draw neater paths when the number of children is 2 or more. However, I can't make this work properly as I can't get things to expand in the right places at the right times. So this is cruder and more fragile. (That is, it is uglier and will even more likely break.)
Caveat emptor...
\documentclass[tikz, border=5pt]{standalone}
\usepackage{forest}
\usetikzlibrary{arrows.meta}
\begin{document}
\tikzset{
enclosure/.style={
draw=blue,
}
}
\begin{forest}
for tree={
circle,
draw,
outer sep=2pt,
s sep+=2.5pt,
edge={shorten <=-2pt, shorten >=-2pt},
if level=0{
for children={
if n=1{
tikz={
\draw [enclosure] (.west) [out=90, in=-135] to (.north west) -- (!u.north west);
\draw [enclosure] (.east) -- (!n.north west |- !u.south) -- (!n.north west) [out=-135, in=90] to (!n.west);
},
}{
if n'=1{
tikz={
\draw [enclosure, -{Stealth[]}] (.east) [out=90, in=-45] to (.north east) -- (!u.north east) ;
\draw [enclosure] (.west) -- (!p.north east |- !u.south) -- (!p.north east) [out=-45, in=90] to (!p.east);
},
}{}
}
},
}{
if n children=1{
for children={
tikz={
\draw [enclosure] (.west) -- (!u.west) [out=90, in=-135] to (!u.north west) (.east) -- (!u.east) [out=90, in=-45] to (!u.north east) ;
},
}
}{
if n children=2{
for children={
if n=1{
tikz={
\draw [enclosure] (.west) [out=90, in=-135] to (.north west) -- (!u.west) (.east) [out=90, in=-45] to (.north east) -- (!u.south);
},
}{
if n'=1{
tikz={
\draw [enclosure] (.east) [out=90, in=-45] to (.north east) -- (!u.east) (.west) [out=90, in=-135] to (.north west) -- (!u.south);
},
}{}
}
},
}{
for children={
if n=1{
tikz={
\draw [enclosure] (.west) [out=90, in=-135] to (.north west) -- (!u.west);
\draw [enclosure] (.east) -- (!n.north west |- !u.south) -- (!n.north west) [out=-135, in=90] to (!n.west);
},
}{
if n'=1{
tikz={
\draw [enclosure] (.east) [out=90, in=-45] to (.north east) -- (!u.east);
\draw [enclosure] (.west) -- (!p.north east |- !u.south) -- (!p.north east) [out=-45, in=90] to (!p.east);
},
}{}
}
},
}
},
if n children=0{
afterthought={
\draw [enclosure] (.west) [out=-90, in=-180] to (.south) [out=0, in=-90] to (.east);
}
}{}
},
if={isodd(n_children)}{
for children={
if={equal(n,int((n_children("!u")+1)/2))}{
calign with current
}{}
}
}{},
}
[A
[B
[E]
[F
[K]
[L]
[M]
]
[G]
]
[C
[H]
]
[D
[I]
[J
[N]
[O]
]
]
]
\end{forest}
\end{document}
cfr
- 198,882