我怎样才能实现二叉树深度优先遍历的绘制这张照片？
答案1
\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}
答案2
这个解决方案不如手绘版本那么漂亮。但是，它在某种程度上是自动化的。话虽如此，如果一棵树与我测试的那棵树差别太大，它肯定会产生不想要的结果。
解决方案使用forest（当然）。我想使用intersections库来获取交叉点，以便在子项数量为 2 或更多时绘制更整洁的路径。但是，我无法使其正常工作，因为我无法让事物在正确的时间在正确的位置扩展。所以这更粗糙，更脆弱。（也就是说，它更丑陋，甚至更容易损坏。）
买者自负...
\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}