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Below is what picture i want enter image description here

Words of level 1: Rule

Words of level 2: Length Pattern Arcs

Words of level 3: T1 T2 A1

Words of level 4: term_index term pos_tag ne_tag term_index term pos_tag ne_tag start_index end_index semantic_type

String of bottom line: 2 [1][][][LOC] [2][][][LOC] -> 1-2:loc-cons

Below is my current code, and the picture generated by it.




\begin{tikzpicture}[edge from parent fork down]

\tikzstyle{level 2}=[level distance=1.3cm, sibling distance=10cm]
\tikzstyle{level 3}=[level distance=2.0cm, sibling distance=5cm]
\tikzstyle{level 3}=[level distance=2.0cm, sibling distance=3cm]

\node {Rule}
    child {node {Length}
        child {node {2}}
    child {node {Pattern}
        child {node {$T_1$}
            child {node[r] {term\_index}}
            child {node[r] {term}}
            child {node[r] {pos\_tag}}
            child {node[r] {ne\_tag}}
        child {node {$T_2$}
            child {node[r] {term\_index}}
            child {node[r] {term}}
            child {node[r] {pos\_tag}}
            child {node[r] {ne\_tag}}
    child {node {Arcs}
        child {node {$A_1$}
            child {node[r] {start\_index}}
            child {node[r] {end\_index}}
            child {node[r] {semantic\_type}}

%    \node{2[][][][LOC] [2][][][LOC] \rightarrow 1-2:loc-cons};



enter image description here

Can anyone help me with this, thanks a lot!

share|improve this question
I would use tikz-qtree for this. Here are some related questions: How to draw up a hierarchical tree diagram for taxonomic classification and Joining two branches of a family tree. – Alan Munn May 30 '12 at 16:42
@AlanMunn thanks, i will try it – icycandy May 30 '12 at 16:52
up vote 3 down vote accepted

As you're stuck with the design of the bottom line, I thought it was easier to build the tree from this bottom line without using a tree-package but instead of using a pre-defined distance between levels and barycentric coordinates system.

I placed the labels of the bottom line below instead of above because then it seems clearer to me.

\usetikzlibrary{calc, positioning} 


% place bottom line
\node (n00) {2};
\foreach \str/\l/\comment [count=\num]in {%
    \node[right=\l of n0\leftnum] (n0\num) {\str};
    \node[rotate=45,below left=0mm of n0\num.center] (c\num) {\textcolor{gray}{\sffamily\comment}};

% distance between two levels

% level 1
\node (n10) at ($(n00)+(0,\distance)$) {};
\node (n11) at ($(barycentric cs:n01=1,n02=1,n03=1,n04=1)+(0,\distance)$) {$T_1$};
\node (n12) at ($(barycentric cs:n05=1,n06=1,n07=1,n08=1)+(0,\distance)$) {$T_2$};
\node (n13) at ($(barycentric cs:n09=1,n010=1,n011=1)+(0,\distance)$) {$A_1$};

% level 2
\node (n20) at ($(n10)+(0,\distance)$) {Length};
\node (n21) at ($(barycentric cs:n11=1,n12=1)+(0,\distance)$) {Pattern};
\node (n22) at ($(n13)+(0,\distance)$) {Arcs};

% level 3
\node (n30) at ($(barycentric cs:n20=1,n21=1,n22=1)+(0,\distance)$) {Rule};

% links
\draw (n00) -- (n20);
\foreach \num in {1,...,4}
    \draw ($(n0\num.north)+(0,1mm)$) |- ($(n11)+(0,-1)$);
\draw ($(n11)+(0,-1)$) -- (n11);
\foreach \num in {5,...,8}
    \draw ($(n0\num.north)+(0,1mm)$) |- ($(n12)+(0,-1)$);
\draw ($(n12)+(0,-1)$) -- (n12);
\foreach \num in {9,...,11}
    \draw ($(n0\num.north)+(0,1mm)$) |- ($(n13)+(0,-1)$);
\draw ($(n13)+(0,-1)$) -- (n13);
\draw (n11) |- ($(n21)+(0,-1)$) -- (n21);
\draw (n12) |- ($(n21)+(0,-1)$);
\draw (n13) -- (n22);
\foreach \num in {0,...,2}
    \draw ($(n2\num.north)+(0,1mm)$) |- ($(n30)+(0,-1)$);
\draw ($(n30)+(0,-1)$) -- (n30);


Here is what the code above produces:

the tree

share|improve this answer
I'm doing my master thesis, as I'm not familiar with LaTeX, your answer help me a lot. Thank you! – icycandy Jun 4 '12 at 16:49

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