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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。）

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自动封闭

\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}

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