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# Abstract Recursive Trees

The purpose of this class is to help implement trees with a specific structure on the children of each node. For instance, one could want to define a tree in which each node sees its children as linearly (see the `Ordered Trees` module) or cyclically ordered.

Tree structures

Conceptually, one can define a tree structure from any object that can contain others. Indeed, a list can contain lists which contain lists which contain lists, and thus define a tree ... The same can be done with sets, or any kind of iterable objects.

While any iterable is sufficient to encode trees, it can prove useful to have other methods available like isomorphism tests (see next section), conversions to DiGraphs objects (see `as_digraph()`) or computation of the number of automorphisms constrained by the structure on children. Providing such methods is the whole purpose of the `AbstractTree` class.

As a result, the `AbstractTree` class is not meant to be instantiated, but extended. It is expected that classes extending this one may also inherit from classes representing iterables, for instance `ClonableArray` or `ClonableList`

Constrained Trees

The tree built from a specific container will reflect the properties of the container. Indeed, if `A` is an iterable class whose elements are linearly ordered, a class `B` extending both of `AbstractTree` and `A` will be such that the children of a node will be linearly ordered. If `A` behaves like a set (i.e. if there is no order on the elements it contains), then two trees will be considered as equal if one can be obtained from the other through permutations between the children of a same node (see next section).

Paths and ID

It is expected that each element of a set of children should be identified by its index in the container. This way, any node of the tree can be identified by a word describing a path from the root node.

Canonical labellings

Equality between instances of classes extending both `AbstractTree` and `A` is entirely defined by the equality defined on the elements of `A`. A canonical labelling of such a tree, however, should be such that two trees `a` and `b` satisfying `a == b` have the same canonical labellings. On the other hand, the canonical labellings of trees `a` and `b` satisfying `a != b` are expected to be different.

For this reason, the values returned by the `canonical_labelling` method heavily depend on the data structure used for a node’s children and should be overridden by most of the classes extending `AbstractTree` if it is incoherent with the data structure.

Authors

- Florent Hivert (2010-2011): initial revision
- Frédéric Chapoton (2011): contributed some methods

class `sage.combinat.abstract_tree.``AbstractClonableTree`

  Bases: `sage.combinat.abstract_tree.AbstractTree`

  Abstract Clonable Tree.

  An abstract class for trees with clone protocol (see `list_clone`). It is expected that classes extending this one may also inherit from classes like `ClonableArray` or `ClonableList` depending whether one wants to build trees where adding a child is allowed.

  Note

  Due to the limitation of Cython inheritance, one cannot inherit here from `ClonableElement`, because it would prevent us from later inheriting from `ClonableArray` or `ClonableList`.

  How should this class be extended ?

  A class extending `AbstractClonableTree` should satisfy the following assumptions:

  - An instantiable class extending `AbstractClonableTree` should also extend the `ClonableElement` class or one of its subclasses generally, at least `ClonableArray`.
  - To respect the Clone protocol, the `AbstractClonableTree.check()` method should be overridden by the new class.

  See also the assumptions in `AbstractTree`.

  `check`()

    Check that `self` is a correct tree.

    This method does nothing. It is implemented here because many extensions of `AbstractClonableTree` also extend `sage.structure.list_clone.ClonableElement`, which requires it.

    It should be overridden in subclasses in order to check that the characterizing property of the respective kind of tree holds (eg: two children for binary trees).

    EXAMPLES:

        sage: OrderedTree([[],[[]]]).check()
        sage: BinaryTree([[],[[],[]]]).check()

class `sage.combinat.abstract_tree.``AbstractLabelledClonableTree`(parent, children, label=None, check=True)

  Bases: `sage.combinat.abstract_tree.AbstractLabelledTree`, `sage.combinat.abstract_tree.AbstractClonableTree`

  Abstract Labelled Clonable Tree

  This class takes care of modification for the label by the clone protocol.

  Note

  Due to the limitation of Cython inheritance, one cannot inherit here from `ClonableArray`, because it would prevent us to inherit later from `ClonableList`.

  `map_labels`(f)

    Applies the function \(f\) to the labels of `self`

    This method returns a copy of `self` on which the function \(f\) has been applied on all labels (a label \(x\) is replaced by \(f(x)\)).

    EXAMPLES:

        sage: LT = LabelledOrderedTree
        sage: t = LT([LT([],label=1),LT([],label=7)],label=3); t
        3[1[], 7[]]
        sage: t.map_labels(lambda z:z+1)
        4[2[], 8[]]

        sage: LBT = LabelledBinaryTree
        sage: bt = LBT([LBT([],label=1),LBT([],label=4)],label=2); bt
        2[1[., .], 4[., .]]
        sage: bt.map_labels(lambda z:z+1)
        3[2[., .], 5[., .]]

  `set_label`(path, label)

    Changes the label of subtree indexed by `path` to `label`

    INPUT:

    - `path` – `None` (default) or a path (list or tuple of children

        index in the tree)

    - `label` – any sage object

    OUPUT: Nothing, `self` is modified in place

    Note

    `self` must be in a mutable state. See `sage.structure.list_clone` for more details about mutability.

    EXAMPLES:

        sage: t = LabelledOrderedTree([[],[[],[]]])
        sage: t.set_label((0,), 4)
        Traceback (most recent call last):
        ...
        ValueError: object is immutable; please change a copy instead.
        sage: with t.clone() as t:
        ....:     t.set_label((0,), 4)
        sage: t
        None[4[], None[None[], None[]]]
        sage: with t.clone() as t:
        ....:     t.set_label((1,0), label = 42)
        sage: t
        None[4[], None[42[], None[]]]

    Todo

    Do we want to implement the following syntactic sugar:

        with t.clone() as tt:
            tt.labels[1,2] = 3 ?

  `set_root_label`(label)

    Sets the label of the root of `self`

    INPUT: `label` – any Sage object

    OUPUT: `None`, `self` is modified in place

    Note

    `self` must be in a mutable state. See `sage.structure.list_clone` for more details about mutability.

    EXAMPLES:

        sage: t = LabelledOrderedTree([[],[[],[]]])
        sage: t.set_root_label(3)
        Traceback (most recent call last):
        ...
        ValueError: object is immutable; please change a copy instead.
        sage: with t.clone() as t:
        ....:     t.set_root_label(3)
        sage: t.label()
        3
        sage: t
        3[None[], None[None[], None[]]]

    This also works for binary trees:

        sage: bt = LabelledBinaryTree([[],[]])
        sage: bt.set_root_label(3)
        Traceback (most recent call last):
        ...
        ValueError: object is immutable; please change a copy instead.
        sage: with bt.clone() as bt:
        ....:     bt.set_root_label(3)
        sage: bt.label()
        3
        sage: bt
        3[None[., .], None[., .]]

    TESTS:

        sage: with t.clone() as t:
        ....:     t[0] = LabelledOrderedTree(t[0], label = 4)
        sage: t
        3[4[], None[None[], None[]]]
        sage: with t.clone() as t:
        ....:     t[1,0] = LabelledOrderedTree(t[1,0], label = 42)
        sage: t
        3[4[], None[42[], None[]]]

class `sage.combinat.abstract_tree.``AbstractLabelledTree`(parent, children, label=None, check=True)

  Bases: `sage.combinat.abstract_tree.AbstractTree`

  Abstract Labelled Tree.

  Typically a class for labelled trees is contructed by inheriting from a class for unlabelled trees and `AbstractLabelledTree`.

  How should this class be extended ?

  A class extending `AbstractLabelledTree` should respect the following assumptions:

  - For a labelled tree `T` the call `T.parent().unlabelled_trees()` should return a parent for unlabelled trees of the same kind: for example,
    - if `T` is a binary labelled tree, `T.parent()` is `LabelledBinaryTrees()` and `T.parent().unlabelled_trees()` is `BinaryTrees()`
    - if `T` is an ordered labelled tree, `T.parent()` is `LabelledOrderedTrees()` and `T.parent().unlabelled_trees()` is `OrderedTrees()`
  - In the same vein, the class of `T` should contain an attribute `_UnLabelled` which should be the class for the corresponding unlabelled trees.

  See also the assumptions in `AbstractTree`.

  See also

  `AbstractTree`

  `as_digraph`()

    Returns a directed graph version of `self`.

    Warning

    At this time, the output makes sense only if `self` is a labelled binary tree with no repeated labels and no `None` labels.

    EXAMPLES:

        sage: LT = LabelledOrderedTrees()
        sage: t1 = LT([LT([],label=6),LT([],label=1)],label=9)
        sage: t1.as_digraph()
        Digraph on 3 vertices

        sage: t = BinaryTree([[None, None],[[],None]]);
        sage: lt = t.canonical_labelling()
        sage: lt.as_digraph()
        Digraph on 4 vertices

  `label`(path=None)

    Return the label of `self`.

    INPUT:

    - `path` – None (default) or a path (list or tuple of children index

        in the tree)

    OUTPUT: the label of the subtree indexed by `path`

    EXAMPLES:

        sage: t = LabelledOrderedTree([[],[]], label = 3)
        sage: t.label()
        3
        sage: t[0].label()
        sage: t = LabelledOrderedTree([LabelledOrderedTree([], 5),[]], label = 3)
        sage: t.label()
        3
        sage: t[0].label()
        5
        sage: t[1].label()
        sage: t.label([0])
        5

  `labels`()

    Return the list of labels of `self`.

    EXAMPLES:

        sage: LT = LabelledOrderedTree
        sage: t = LT([LT([],label='b'),LT([],label='c')],label='a')
        sage: t.labels()
        ['a', 'b', 'c']

        sage: LBT = LabelledBinaryTree
        sage: LBT([LBT([],label=1),LBT([],label=4)],label=2).labels()
        [2, 1, 4]

  `leaf_labels`()

    Return the list of labels of the leaves of `self`.

    In case of a labelled binary tree, these “leaves” are not actually the leaves of the binary trees, but the nodes whose both children are leaves!

    EXAMPLES:

        sage: LT = LabelledOrderedTree
        sage: t = LT([LT([],label='b'),LT([],label='c')],label='a')
        sage: t.leaf_labels()
        ['b', 'c']

        sage: LBT = LabelledBinaryTree
        sage: bt = LBT([LBT([],label='b'),LBT([],label='c')],label='a')
        sage: bt.leaf_labels()
        ['b', 'c']
        sage: LBT([], label='1').leaf_labels()
        ['1']
        sage: LBT(None).leaf_labels()
        []

  `shape`()

    Returns the unlabelled tree associated to `self`

    EXAMPLES:

        sage: t = LabelledOrderedTree([[],[[]]], label = 25).shape(); t
        [[], [[]]]

        sage: LabelledBinaryTree([[],[[],[]]], label = 25).shape()
        [[., .], [[., .], [., .]]]

    TESTS:

        sage: t.parent()
        Ordered trees
        sage: type(t)
        <class 'sage.combinat.ordered_tree.OrderedTrees_all_with_category.element_class'>

class `sage.combinat.abstract_tree.``AbstractTree`

  Bases: `object`

  Abstract Tree.

  There is no data structure defined here, as this class is meant to be extended, not instantiated.

  How should this class be extended?

  A class extending `AbstractTree` should respect several assumptions:

  - For a tree `T`, the call `iter(T)` should return an iterator on the children of the root `T`.
  - The `canonical_labelling` method should return the same value for trees that are considered equal (see the “canonical labellings” section in the documentation of the `AbstractTree` class).
  - For a tree `T` the call `T.parent().labelled_trees()` should return a parent for labelled trees of the same kind: for example,
    - if `T` is a binary tree, `T.parent()` is `BinaryTrees()` and `T.parent().labelled_trees()` is `LabelledBinaryTrees()`
    - if `T` is an ordered tree, `T.parent()` is `OrderedTrees()` and `T.parent().labelled_trees()` is `LabelledOrderedTrees()`

  TESTS:

      sage: TestSuite(OrderedTree()).run()
      sage: TestSuite(BinaryTree()).run()

  `breadth_first_order_traversal`(action=None)

    Run the breadth-first post-order traversal algorithm and subject every node encountered to some procedure `action`. The algorithm is:

        queue <- [ root ];
        while the queue is not empty:
            node <- pop( queue );
            manipulate the node;
            prepend to the queue the list of all subtrees of
                the node (from the rightmost to the leftmost).

    INPUT:

    - `action` – (optional) a function which takes a node as input, and does something during the exploration

    OUTPUT:

    `None`. (This is not an iterator.)

    EXAMPLES:

    For example, on the following binary tree \(b\):

        |   ___3____      |
        |  /        \     |
        | 1         _7_   |
        |  \       /   \  |
        |   2     5     8 |
        |        / \      |
        |       4   6     |

    the breadth-first order traversal algorithm explores \(b\) in the following order of nodes: \(3,1,7,2,5,8,4,6\).

    TESTS:

        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: l = []
        sage: b.breadth_first_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [3, 1, 7, 2, 5, 8, 4, 6]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: t
        1[2[3[4[], 5[]]], 6[7[]], 8[9[], 10[]]]
        sage: l = []
        sage: t.breadth_first_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1, 2, 6, 8, 3, 7, 9, 10, 4, 5]

        sage: l = []
        sage: BinaryTree().canonical_labelling().\
        ....:     breadth_first_order_traversal(
        ....:         lambda node: l.append(node.label()))
        sage: l
        []
        sage: OrderedTree([]).canonical_labelling().\
        ....:     breadth_first_order_traversal(
        ....:         lambda node: l.append(node.label()))
        sage: l
        [1]

  `canonical_labelling`(shift=1)

    Returns a labelled version of `self`.

    The actual canonical labelling is currently unspecified. However, it is guaranteed to have labels in \(1...n\) where \(n\) is the number of nodes of the tree. Moreover, two (unlabelled) trees compare as equal if and only if their canonical labelled trees compare as equal.

    EXAMPLES:

        sage: t = OrderedTree([[], [[], [[], []], [[], []]], [[], []]])
        sage: t.canonical_labelling()
        1[2[], 3[4[], 5[6[], 7[]], 8[9[], 10[]]], 11[12[], 13[]]]

        sage: BinaryTree([]).canonical_labelling()
        1[., .]
        sage: BinaryTree([[],[[],[]]]).canonical_labelling()
        2[1[., .], 4[3[., .], 5[., .]]]

    TESTS:

        sage: BinaryTree().canonical_labelling()
        .

  `depth`()

    The depth of `self`.

    EXAMPLES:

        sage: OrderedTree().depth()
        1
        sage: OrderedTree([]).depth()
        1
        sage: OrderedTree([[],[]]).depth()
        2
        sage: OrderedTree([[],[[]]]).depth()
        3
        sage: OrderedTree([[], [[], [[], []], [[], []]], [[], []]]).depth()
        4

        sage: BinaryTree().depth()
        0
        sage: BinaryTree([[],[[],[]]]).depth()
        3

  `iterative_post_order_traversal`(action=None)

    Run the depth-first post-order traversal algorithm (iterative implementation) and subject every node encountered to some procedure `action`. The algorithm is:

        explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one;
        then manipulate the root with function `action` (in the
            case of a binary tree, only if the root is not a leaf).

    INPUT:

    - `action` – (optional) a function which takes a node as input, and does something during the exploration

    OUTPUT:

    `None`. (This is not an iterator.)

    See also

    - `post_order_traversal_iter()`

    TESTS:

        sage: l = []
        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: b
        3[1[., 2[., .]], 7[5[4[., .], 6[., .]], 8[., .]]]
        sage: b.iterative_post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [2, 1, 4, 6, 5, 8, 7, 3]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: t
        1[2[3[4[], 5[]]], 6[7[]], 8[9[], 10[]]]
        sage: l = []
        sage: t.iterative_post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [4, 5, 3, 2, 7, 6, 9, 10, 8, 1]

        sage: l = []
        sage: BinaryTree().canonical_labelling().\
        ....:     iterative_post_order_traversal(
        ....:         lambda node: l.append(node.label()))
        sage: l
        []
        sage: OrderedTree([]).canonical_labelling().\
        ....:     iterative_post_order_traversal(
        ....:         lambda node: l.append(node.label()))
        sage: l
        [1]

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = []
        sage: t.iterative_post_order_traversal(lambda node: l.append(1))
        sage: len(l)
        7

  `iterative_pre_order_traversal`(action=None)

    Run the depth-first pre-order traversal algorithm (iterative implementation) and subject every node encountered to some procedure `action`. The algorithm is:

        manipulate the root with function `action` (in the case
            of a binary tree, only if the root is not a leaf);
        then explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one.

    INPUT:

    - `action` – (optional) a function which takes a node as input, and does something during the exploration

    OUTPUT:

    `None`. (This is not an iterator.)

    See also

    - `pre_order_traversal_iter()`
    - `pre_order_traversal()`

    TESTS:

        sage: l = []
        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: b
        3[1[., 2[., .]], 7[5[4[., .], 6[., .]], 8[., .]]]
        sage: b.iterative_pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [3, 1, 2, 7, 5, 4, 6, 8]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: t
        1[2[3[4[], 5[]]], 6[7[]], 8[9[], 10[]]]
        sage: l = []
        sage: t.iterative_pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
        sage: l = []

        sage: BinaryTree().canonical_labelling().\
        ....:     pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        []
        sage: OrderedTree([]).canonical_labelling().\
        ....:     iterative_pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1]

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = []
        sage: t.iterative_pre_order_traversal(lambda node: l.append(1))
        sage: len(l)
        7

  `node_number`()

    The number of nodes of `self`.

    EXAMPLES:

        sage: OrderedTree().node_number()
        1
        sage: OrderedTree([]).node_number()
        1
        sage: OrderedTree([[],[]]).node_number()
        3
        sage: OrderedTree([[],[[]]]).node_number()
        4
        sage: OrderedTree([[], [[], [[], []], [[], []]], [[], []]]).node_number()
        13

    EXAMPLES:

        sage: BinaryTree(None).node_number()
        0
        sage: BinaryTree([]).node_number()
        1
        sage: BinaryTree([[], None]).node_number()
        2
        sage: BinaryTree([[None, [[], []]], None]).node_number()
        5

  `paths`()

    Return a generator for all paths to nodes of `self`.

    OUTPUT:

    This method returns a list of sequences of integers. Each of these sequences represents a path from the root node to some node. For instance, \((1, 3, 2, 5, 0, 3)\) represents the node obtained by choosing the 1st child of the root node (in the ordering returned by `iter`), then the 3rd child of its child, then the 2nd child of the latter, etc. (where the labelling of the children is zero-based).

    The root element is represented by the empty tuple `()`.

    EXAMPLES:

        sage: list(OrderedTree([]).paths())
        [()]
        sage: list(OrderedTree([[],[[]]]).paths())
        [(), (0,), (1,), (1, 0)]

        sage: list(BinaryTree([[],[[],[]]]).paths())
        [(), (0,), (1,), (1, 0), (1, 1)]

    TESTS:

        sage: t = OrderedTree([[], [[], [[], []], [[], []]], [[], []]])
        sage: t.node_number() == len(list(t.paths()))
        True
        sage: list(BinaryTree().paths())
        []
        sage: bt = BinaryTree([[],[[],[]]])
        sage: bt.node_number() == len(list(bt.paths()))
        True

  `post_order_traversal`(action=None)

    Run the depth-first post-order traversal algorithm (recursive implementation) and subject every node encountered to some procedure `action`. The algorithm is:

        explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one;
        then manipulate the root with function `action` (in the
            case of a binary tree, only if the root is not a leaf).

    INPUT:

    - `action` – (optional) a function which takes a node as input, and does something during the exploration

    OUTPUT:

    `None`. (This is not an iterator.)

    See also

    - `post_order_traversal_iter()`
    - `iterative_post_order_traversal()`

    TESTS:

        sage: l = []
        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: b
        3[1[., 2[., .]], 7[5[4[., .], 6[., .]], 8[., .]]]
        sage: b.post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [2, 1, 4, 6, 5, 8, 7, 3]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).\
        ....:     canonical_labelling(); t
        1[2[3[4[], 5[]]], 6[7[]], 8[9[], 10[]]]
        sage: l = []
        sage: t.post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [4, 5, 3, 2, 7, 6, 9, 10, 8, 1]

        sage: l = []
        sage: BinaryTree().canonical_labelling().\
        ....:     post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        []
        sage: OrderedTree([]).canonical_labelling().\
        ....:     post_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1]

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = []
        sage: t.post_order_traversal(lambda node: l.append(1))
        sage: len(l)
        7

  `post_order_traversal_iter`()

    The depth-first post-order traversal iterator.

    This method iters each node following the depth-first post-order traversal algorithm (recursive implementation). The algorithm is:

        explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one;
        then yield the root (in the case of binary trees, only if
            it is not a leaf).

    EXAMPLES:

    For example on the following binary tree \(b\):

        |   ___3____      |
        |  /        \     |
        | 1         _7_   |
        |  \       /   \  |
        |   2     5     8 |
        |        / \      |
        |       4   6     |

    (only the nodes are shown), the depth-first post-order traversal algorithm explores \(b\) in the following order of nodes: \(2,1,4,6,5,8,7,3\).

    For another example, consider the labelled tree:

        |     __1____ |
        |    /  /   / |
        |   2  6   8_ |
        |   |  |  / / |
        |   3_ 7 9 10 |
        |  / /        |
        | 4 5         |

    The algorithm explores this tree in the following order: \(4,5,3,2,7,6,9,10,8,1\).

    TESTS:

        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: ascii_art([b])
        [   ___3____      ]
        [  /        \     ]
        [ 1         _7_   ]
        [  \       /   \  ]
        [   2     5     8 ]
        [        / \      ]
        [       4   6     ]
        sage: [node.label() for node in b.post_order_traversal_iter()]
        [2, 1, 4, 6, 5, 8, 7, 3]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: ascii_art([t])
        [     __1____   ]
        [    /  /   /   ]
        [   2  6   8_   ]
        [   |  |  / /   ]
        [   3_ 7 9 10   ]
        [  / /          ]
        [ 4 5           ]
        sage: [node.label() for node in t.post_order_traversal_iter()]
        [4, 5, 3, 2, 7, 6, 9, 10, 8, 1]

        sage: [node.label() for node in BinaryTree().canonical_labelling().\
        ....:     post_order_traversal_iter()]
        []
        sage: [node.label() for node in OrderedTree([]).\
        ....:     canonical_labelling().post_order_traversal_iter()]
        [1]

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = [1 for i in t.post_order_traversal_iter()]
        sage: len(l)
        7

  `pre_order_traversal`(action=None)

    Run the depth-first pre-order traversal algorithm (recursive implementation) and subject every node encountered to some procedure `action`. The algorithm is:

        manipulate the root with function `action` (in the case
            of a binary tree, only if the root is not a leaf);
        then explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one.

    INPUT:

    - `action` – (optional) a function which takes a node as input, and does something during the exploration

    OUTPUT:

    `None`. (This is not an iterator.)

    EXAMPLES:

    For example, on the following binary tree \(b\):

        |   ___3____      |
        |  /        \     |
        | 1         _7_   |
        |  \       /   \  |
        |   2     5     8 |
        |        / \      |
        |       4   6     |

    the depth-first pre-order traversal algorithm explores \(b\) in the following order of nodes: \(3,1,2,7,5,4,6,8\).

    Another example:

        |     __1____ |
        |    /  /   / |
        |   2  6   8_ |
        |   |  |  / / |
        |   3_ 7 9 10 |
        |  / /        |
        | 4 5         |

    The algorithm explores this tree in the following order: \(1,2,3,4,5,6,7,8,9,10\).

    See also

    - `pre_order_traversal_iter()`
    - `iterative_pre_order_traversal()`

    TESTS:

        sage: l = []
        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: b
        3[1[., 2[., .]], 7[5[4[., .], 6[., .]], 8[., .]]]
        sage: b.pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [3, 1, 2, 7, 5, 4, 6, 8]
        sage: li = []
        sage: b.iterative_pre_order_traversal(lambda node: li.append(node.label()))
        sage: l == li
        True

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: t
        1[2[3[4[], 5[]]], 6[7[]], 8[9[], 10[]]]
        sage: l = []
        sage: t.pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
        sage: li = []
        sage: t.iterative_pre_order_traversal(lambda node: li.append(node.label()))
        sage: l == li
        True

        sage: l = []
        sage: BinaryTree().canonical_labelling().\
        ....:     pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        []
        sage: OrderedTree([]).canonical_labelling().\
        ....:     pre_order_traversal(lambda node: l.append(node.label()))
        sage: l
        [1]

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = []
        sage: t.pre_order_traversal(lambda node: l.append(1))
        sage: len(l)
        7

  `pre_order_traversal_iter`()

    The depth-first pre-order traversal iterator.

    This method iters each node following the depth-first pre-order traversal algorithm (recursive implementation). The algorithm is:

        yield the root (in the case of binary trees, if it is not
            a leaf);
        then explore each subtree (by the algorithm) from the
            leftmost one to the rightmost one.

    EXAMPLES:

    For example, on the following binary tree \(b\):

        |   ___3____      |
        |  /        \     |
        | 1         _7_   |
        |  \       /   \  |
        |   2     5     8 |
        |        / \      |
        |       4   6     |

    (only the nodes shown), the depth-first pre-order traversal algorithm explores \(b\) in the following order of nodes: \(3,1,2,7,5,4,6,8\).

    Another example:

        |     __1____ |
        |    /  /   / |
        |   2  6   8_ |
        |   |  |  / / |
        |   3_ 7 9 10 |
        |  / /        |
        | 4 5         |

    The algorithm explores this labelled tree in the following order: \(1,2,3,4,5,6,7,8,9,10\).

    TESTS:

        sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
        sage: ascii_art([b])
        [   ___3____      ]
        [  /        \     ]
        [ 1         _7_   ]
        [  \       /   \  ]
        [   2     5     8 ]
        [        / \      ]
        [       4   6     ]
        sage: [n.label() for n in b.pre_order_traversal_iter()]
        [3, 1, 2, 7, 5, 4, 6, 8]

        sage: t = OrderedTree([[[[],[]]],[[]],[[],[]]]).canonical_labelling()
        sage: ascii_art([t])
        [     __1____   ]
        [    /  /   /   ]
        [   2  6   8_   ]
        [   |  |  / /   ]
        [   3_ 7 9 10   ]
        [  / /          ]
        [ 4 5           ]
        sage: [n.label() for n in t.pre_order_traversal_iter()]
        [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

        sage: [n for n in BinaryTree(None).pre_order_traversal_iter()]
        []

    The following test checks that things do not go wrong if some among the descendants of the tree are equal or even identical:

        sage: u = BinaryTree(None)
        sage: v = BinaryTree([u, u])
        sage: w = BinaryTree([v, v])
        sage: t = BinaryTree([w, w])
        sage: t.node_number()
        7
        sage: l = [1 for i in t.pre_order_traversal_iter()]
        sage: len(l)
        7

  `subtrees`()

    Return a generator for all nonempty subtrees of `self`.

    The number of nonempty subtrees of a tree is its number of nodes. (The word “nonempty” makes a difference only in the case of binary trees. For ordered trees, for example, all trees are nonempty.)

    EXAMPLES:

        sage: list(OrderedTree([]).subtrees())
        [[]]
        sage: list(OrderedTree([[],[[]]]).subtrees())
        [[[], [[]]], [], [[]], []]

        sage: list(OrderedTree([[],[[]]]).canonical_labelling().subtrees())
        [1[2[], 3[4[]]], 2[], 3[4[]], 4[]]

        sage: list(BinaryTree([[],[[],[]]]).subtrees())
        [[[., .], [[., .], [., .]]], [., .], [[., .], [., .]], [., .], [., .]]

        sage: v = BinaryTree([[],[]])
        sage: list(v.canonical_labelling().subtrees())
        [2[1[., .], 3[., .]], 1[., .], 3[., .]]

    TESTS:

        sage: t = OrderedTree([[], [[], [[], []], [[], []]], [[], []]])
        sage: t.node_number() == len(list(t.subtrees()))
        True
        sage: list(BinaryTree().subtrees())
        []
        sage: bt = BinaryTree([[],[[],[]]])
        sage: bt.node_number() == len(list(bt.subtrees()))
        True

  `to_hexacode`()

    Transform a tree into an hexadecimal string.

    The definition of the hexacode is recursive. The first letter is the valence of the root as an hexadecimal (up to 15), followed by the concatenation of the hexacodes of the subtrees.

    This method only works for trees where every vertex has valency at most 15.

    See `from_hexacode()` for the reverse transformation.

    EXAMPLES:

        sage: from sage.combinat.abstract_tree import from_hexacode
        sage: LT = LabelledOrderedTrees()
        sage: from_hexacode('2010', LT).to_hexacode()
        '2010'
        sage: LT.an_element().to_hexacode()
        '3020010'
        sage: t = from_hexacode('a0000000000000000', LT)
        sage: t.to_hexacode()
        'a0000000000'

        sage: OrderedTrees(6).an_element().to_hexacode()
        '500000'

    TESTS:

        sage: one = LT([], label='@')
        sage: LT([one for _ in range(15)], label='@').to_hexacode()
        'f000000000000000'
        sage: LT([one for _ in range(16)], label='@').to_hexacode()
        Traceback (most recent call last):
        ...
        ValueError: the width of the tree is too large

  `tree_factorial`()

    Return the tree-factorial of `self`.

    Definition:

    The tree-factorial \(T!\) of a tree \(T\) is the product \(\prod_{v\in T}\#\mbox{children}(v)\).

    EXAMPLES:

        sage: LT = LabelledOrderedTrees()
        sage: t = LT([LT([],label=6),LT([],label=1)],label=9)
        sage: t.tree_factorial()
        3

        sage: BinaryTree([[],[[],[]]]).tree_factorial()
        15

    TESTS:

        sage: BinaryTree().tree_factorial()
        1

`sage.combinat.abstract_tree.``from_hexacode`(ch, parent=None, label='@')

  Transform an hexadecimal string into a tree.

  INPUT:

  - `ch` – an hexadecimal string
  - `parent` – kind of trees to be produced. If `None`, this will be `LabelledOrderedTrees`
  - `label` – a label (default: `'@'`) to be used for every vertex of the tree

  See `AbstractTree.to_hexacode()` for the description of the encoding

  See `_from_hexacode_aux()` for the actual code

  EXAMPLES:

      sage: from sage.combinat.abstract_tree import from_hexacode
      sage: from_hexacode('12000', LabelledOrderedTrees())
      @[@[@[], @[]]]

      sage: from_hexacode('1200', LabelledOrderedTrees())
      @[@[@[], @[]]]

  It can happen that only a prefix of the word is used:

      sage: from_hexacode('a'+14*'0', LabelledOrderedTrees())
      @[@[], @[], @[], @[], @[], @[], @[], @[], @[], @[]]

  One can choose the label:

      sage: from_hexacode('1200', LabelledOrderedTrees(), label='o')
      o[o[o[], o[]]]

  One can also create other kinds of trees:

      sage: from_hexacode('1200', OrderedTrees())
      [[[], []]]

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