def ll2nr(l):
"""
"""
if len(l) == 0:
return BinaryTree(None)
elif type(l[2]) != type([]) or type(l[1]) != type([]) or len(l) != 3:
raise ValueError("Given list is not a Binary Tree List")
else:
T = BinaryTree(l[0])
lchild = l[1]
rchild = l[2]
if len(lchild) != 0:
T.insertLeftChild(lchild)
if len(rchild) != 0:
T.insertRightChild(rchild)
if conditions(lchild, rchild):
T.setLeftSubtree(ll2nr(lchild))
T.setRightSubtree(ll2nr(rchild))
return T
def ll2nr(l):
''' (list of lists) -> "Binary Tree" class with multiple .____
Convert a binary tree in a list of lists representation into a nodes
and references form, where each .____ is assigned using the BinaryTree class.
A short explanation: By taking 'right path' or 'left path' what I mean is
passing in that value again into this function, except now since we're looking
at a tree, we will eventually reduce to a single list of elements
(which are not lists). And then that will go to just values which are None or
something else. By saying None, we cover the part where it is a list or it is
something else. At that point, I just put in the appropriate values from the
class.
Note: Here [] means there is nothing there, equivalent to None. The lecture
slides showed None but some students said they would be represented by [], so
I just put both for now.
'''
# Real base case: turning it into a Binary Tree. This is on default because
# we have to come here.
# Initially convert the very first root to a binary tree of a list. It doesn't
# matter about the other elements because the first must always be the root
# node, and therefore holds a value for sure.
greaterTree = BinaryTree(l[0])
# Below are the different 'secondary' base cases:
# If both are nodes, then pass in both at the same time.
if l[1] not in empty_tree and l[2] not in empty_tree:
greaterTree.left = ll2nr(l[1])
greaterTree.right = ll2nr(l[2])
# If the left is a node and the right is not, then take the right path.
elif l[1] in empty_tree and l[2] not in empty_tree:
greaterTree.right = ll2nr(l[2])
# If the left is not a node and the right is, then take the left path.
elif l[1] not in empty_tree and l[2] in empty_tree:
greaterTree.left = ll2nr(l[1])
# If both sides are empty trees, then we give the root. This will slowly trace
# back to the above statement, giving them 'root' values throughout.
elif l[1] in empty_tree and l[2] in empty_tree:
return greaterTree.getRootValue()
# Final output - literally returning instance because no rep method in class.
return greaterTree
def ll2nr (l):
''' (list of lists) -> "Binary Tree" class with multiple .____
Convert a binary tree in a list of lists representation into a nodes
and references form, where each .____ is assigned using the BinaryTree class.
A short explanation: By taking 'right path' or 'left path' what I mean is
passing in that value again into this function, except now since we're looking
at a tree, we will eventually reduce to a single list of elements
(which are not lists). And then that will go to just values which are None or
something else. By saying None, we cover the part where it is a list or it is
something else. At that point, I just put in the appropriate values from the
class.
Note: Here [] means there is nothing there, equivalent to None. The lecture
slides showed None but some students said they would be represented by [], so
I just put both for now.
'''
# Real base case: turning it into a Binary Tree. This is on default because
# we have to come here.
# Initially convert the very first root to a binary tree of a list. It doesn't
# matter about the other elements because the first must always be the root
# node, and therefore holds a value for sure.
greaterTree = BinaryTree(l[0])
# Below are the different 'secondary' base cases:
# If both are nodes, then pass in both at the same time.
if l[1] not in empty_tree and l[2] not in empty_tree:
greaterTree.left = ll2nr(l[1])
greaterTree.right = ll2nr(l[2])
# If the left is a node and the right is not, then take the right path.
elif l[1] in empty_tree and l[2] not in empty_tree:
greaterTree.right = ll2nr(l[2])
# If the left is not a node and the right is, then take the left path.
elif l[1] not in empty_tree and l[2] in empty_tree:
greaterTree.left = ll2nr(l[1])
# If both sides are empty trees, then we give the root. This will slowly trace
# back to the above statement, giving them 'root' values throughout.
elif l[1] in empty_tree and l[2] in empty_tree:
return greaterTree.getRootValue()
# Final output - literally returning instance because no rep method in class.
return greaterTree
