我正在尝试修改我在一个示例中发现的树形方案,但是我无法实现我想要的效果。
我想在图中最后一个矩形下方添加两个矩形。从节点$A_{prv}\Delta P$
应该出现“Q=”这样的行,这为另外两个选项让路。
代码:
\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{trees}
\begin{document}
\tikzstyle{every node}=[draw=black,thick,anchor=west]
\tikzstyle{selected}=[draw=red,fill=red!30]
\tikzstyle{optional}=[dashed,fill=gray!50]
\begin{tikzpicture}[%
grow via three points={one child at (0.5,-0.7) and
two children at (0.5,-0.7) and (0.5,-1.4)},
edge from parent path={(\tikzparentnode.south) |- (\tikzchildnode.west)}]
\node {$(\zeta,\Omega)\rightarrow (y,v_y)$}
child { node {$\dot{\zeta} = \Omega$}}
child { node {$\dot{\Omega} = $}}
child { node {$\dot{P_1} = (B/V)(Q-A\lambda d)$}}
child { node {$\dot{P_2} = (B/V)(Q-A\lambda d)$}}
child { node {Q =}
child { node {$0$ \hspace{14mm} if: \hspace{3mm} $P_1A_1-P_2A_2 < F_{pc}$}}
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}}};
\end{tikzpicture}
\end{document}
答案1
版本 1:最后一个节点的两个子节点。
代替
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}}};
经过
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}
child { node {Child 1 of $A_{prv}\Delta P$}}
child { node {Child 2 of $A_{prv}\Delta P$}}
}
};
\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{trees}
\begin{document}
\tikzstyle{every node}=[draw=black,thick,anchor=west]
\tikzstyle{selected}=[draw=red,fill=red!30]
\tikzstyle{optional}=[dashed,fill=gray!50]
\begin{tikzpicture}[%
grow via three points={one child at (0.5,-0.7) and
two children at (0.5,-0.7) and (0.5,-1.4)},
edge from parent path={(\tikzparentnode.south) |- (\tikzchildnode.west)}]
\node {$(\zeta,\Omega)\rightarrow (y,v_y)$}
child { node {$\dot{\zeta} = \Omega$}}
child { node {$\dot{\Omega} = $}}
child { node {$\dot{P_1} = (B/V)(Q-A\lambda d)$}}
child { node {$\dot{P_2} = (B/V)(Q-A\lambda d)$}}
child { node {Q =}
child { node {$0$ \hspace{14mm} if: \hspace{3mm} $P_1A_1-P_2A_2 < F_{pc}$}}
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}
child { node {Child 1 of $A_{prv}\Delta P$}}
child { node {Child 2 of $A_{prv}\Delta P$}}
}
};
\end{tikzpicture}
\end{document}
版本 2:的另外两个子节点Q =
。
代替
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}}};
经过
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}}
child { node {Child 1 of $A_{prv}\Delta P$}}
child { node {Child 2 of $A_{prv}\Delta P$}}
};
\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{trees}
\begin{document}
\tikzstyle{every node}=[draw=black,thick,anchor=west]
\tikzstyle{selected}=[draw=red,fill=red!30]
\tikzstyle{optional}=[dashed,fill=gray!50]
\begin{tikzpicture}[%
grow via three points={one child at (0.5,-0.7) and
two children at (0.5,-0.7) and (0.5,-1.4)},
edge from parent path={(\tikzparentnode.south) |- (\tikzchildnode.west)}]
\node {$(\zeta,\Omega)\rightarrow (y,v_y)$}
child { node {$\dot{\zeta} = \Omega$}}
child { node {$\dot{\Omega} = $}}
child { node {$\dot{P_1} = (B/V)(Q-A\lambda d)$}}
child { node {$\dot{P_2} = (B/V)(Q-A\lambda d)$}}
child { node {Q =}
child { node {$0$ \hspace{14mm} if: \hspace{3mm} $P_1A_1-P_2A_2 < F_{pc}$}}
child { node {$A_{prv}\Delta P$ \hspace{3mm} if: \hspace{3mm} $P_1A_1-P_2A_2 \geq F_{pc}$}}
child { node {Child 1 of $A_{prv}\Delta P$}}
child { node {Child 2 of $A_{prv}\Delta P$}}
};
\end{tikzpicture}
\end{document}