Function leafCount #
Gives the number of “leaf nodes” in the parse tree of the given expression
A leaf node is one that has no subexpressions, essentially either a
symbol or a constant. Note that 5! has just one leaf, the ‘5’; the
unary factorial operator does not add a leaf. On the other hand,
function symbols do add leaves, so sin(x)/cos(x) has four leaves.
The simplify() function should generally not increase the leafCount()
of an expression, although currently there is no guarantee that it never
does so. In many cases, simplify() reduces the leaf count.
Syntax #
math.leafCount(expr)
Parameters #
| Parameter | Type | Description |
|---|---|---|
expr |
Node | string | The expression to count the leaves of |
Returns #
| Type | Description |
|---|---|
| number | The number of leaves of expr |
Throws #
Type | Description —- | ———–
Examples #
math.leafCount('x') // 1
math.leafCount(math.parse('a*d-b*c')) // 4
math.leafCount('[a,b;c,d][0,1]') // 6
