# 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 #

```
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
```