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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,139 @@ | ||
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| class BinaryTree: | ||
| """ A simple binary tree with left and right branches | ||
| """ | ||
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| def __init__(self, data): | ||
| self._data = data | ||
| self._left = None | ||
| self._right = None | ||
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| def data(self): | ||
| return self._data | ||
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| def left(self): | ||
| return self._left | ||
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| def right(self): | ||
| return self._right | ||
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| def __str__(self): | ||
| """ Returns a pretty string of the tree """ | ||
| def str_branches(node, branches): | ||
| """ Returns a string with the tree pretty printed. | ||
| branches: a list of characters representing the parent branches. Characters can be either ` ` or '│' | ||
| """ | ||
| strings = [str(node._data)] | ||
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| i = 0 | ||
| if node._left != None or node._right != None: | ||
| for current in [node._left, node._right]: | ||
| if i == 0: | ||
| joint = '├' | ||
| else: | ||
| joint = '└' | ||
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| strings.append('\n') | ||
| for b in branches: | ||
| strings.append(b) | ||
| strings.append(joint) | ||
| if i == 0: | ||
| branches.append('│') | ||
| else: | ||
| branches.append(' ') | ||
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| if current != None: | ||
| strings.append(str_branches(current, branches)) | ||
| branches.pop() | ||
| i += 1 | ||
| return "".join(strings) | ||
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| return str_branches(self, []) | ||
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| def insert_left(self, data): | ||
| """ Takes as input DATA (*NOT* a node !!) and MODIFIES current node this way: | ||
| - First creates a new BinaryTree (let's call it B) into which provided data is wrapped. | ||
| - Then: | ||
| - if there is no left node in self, new node B is attached to the left of self | ||
| - if there already is a left node L, it is substituted by new node B, and L becomes the | ||
| left node of B | ||
| """ | ||
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| B = BinaryTree(data) | ||
| if self._left == None: | ||
| self._left = B | ||
| else: | ||
| B._left = self._left | ||
| self._left = B | ||
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| def insert_right(self, data): | ||
| """ Takes as input DATA (*NOT* a node !!) and MODIFIES current node this way: | ||
| - First creates a new BinaryTree (let's call it B) into which provided data is wrapped. | ||
| - Then: | ||
| - if there is no right node in self, new node B is attached to the right of self | ||
| - if there already is a right node L, it is substituted by new node B, and L becomes the | ||
| right node of B | ||
| """ | ||
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| B = BinaryTree(data) | ||
| if self._right == None: | ||
| self._right = B | ||
| else: | ||
| B._right = self._right | ||
| self._right = B | ||
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| def sum_leaves_rec(self): | ||
| """ Supposing the tree holds integer numbers in all nodes, | ||
| RETURN the sum of ONLY the numbers in the leaves. | ||
| - a root with no children is considered a leaf | ||
| - implement it as a recursive Depth First Search (DFS) traversal | ||
| NOTE: with big trees a recursive solution would surely | ||
| exceed the call stack, but here we don't mind | ||
| """ | ||
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| #jupman-raise | ||
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| if self.left() == None and self.right() == None: | ||
| ret = self.data() | ||
| else: | ||
| ret = 0 | ||
| if self.left() != None: | ||
| ret += self.left().sum_leaves_rec() | ||
| if self.right() != None: | ||
| ret += self.right().sum_leaves_rec() | ||
| return ret | ||
| #/jupman-raise | ||
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| def leaves_stack(self): | ||
| """ RETURN a list holding the *data* of all the leaves of the tree, | ||
| in left to right order. | ||
| - a root with no children is considered a leaf | ||
| - DO *NOT* use recursion | ||
| - implement it with a while and a stack (as a python list) | ||
| """ | ||
| #jupman-raise | ||
| ret = [] | ||
| stack = [self] | ||
| while len(stack) > 0: | ||
| node = stack.pop() | ||
| if node.left() == None and node.right() == None: | ||
| ret.append(node.data()) | ||
| # first right then left so we don't need to reverse later | ||
| if node.right() != None: | ||
| stack.append( node.right() ) | ||
| if node.left() != None: | ||
| stack.append( node.left() ) | ||
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| return ret | ||
| #/jupman-raise | ||
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