Forest tree not symmetric

Question

I'm new to the forest package and created a tree with a node between two nodes, that is not centered. In the picture, it's the ZS(A; C) below root. This is the code:

\documentclass{standalone}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage{forest}

\begin{document}
\tikzset{>=latex}

\begin{forest} for tree={align=center}
[TH(3; A; B; C), for children={l+=2ex}
    [TH(2; A; C; B) 
        [TH(1; A; B; C)
            [ZS(A; C)]
        ]
        [ZS(A; B)]
        [TH(1; C; A; B)
            [ZS(C; B)]
        ]
    ] 
    [ZS(A; C)]
    [TH(2; B; A; C)
        [TH(1; B; C; A)
            [ZS(B; A)]
        ]
        [ZS(B; C)]
        [TH(1; A; B; C
            [ZS(A; C)]
        ]
    ]
]
\end{forest}

\end{document}

What am I doing wrong?

Answer 1

This is due to the way forest calculates the node positions, as described on p. 43f of the manual:

For example, consider the manual correction below. By default, B is closer to A than C because packing proceeds from the first to the last child — the position of B would be the same if there was no C. Adjusting s at the right moment, it is easy to center B between A and C.

It goes on to suggest adding before computing xy={s/.average={s}{siblings}} to the non-centred node, like so:

\documentclass{standalone}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage{forest}

\begin{document}
\tikzset{>=latex}

\begin{forest} for tree={align=center}
[TH(3; A; B; C), for children={l+=2ex}
    [TH(2; A; C; B) 
        [TH(1; A; B; C)
            [ZS(A; C)]
        ]
        [ZS(A; B)]
        [TH(1; C; A; B)
            [ZS(C; B)]
        ]
    ] 
    [ZS(A; C), before computing xy={s/.average={s}{siblings}}]
    [TH(2; B; A; C)
        [TH(1; B; C; A)
            [ZS(B; A)]
        ]
        [ZS(B; C)]
        [TH(1; A; B; C
            [ZS(A; C)]
        ]
    ]
]
\end{forest}

\end{document}