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Branches - Julia Bouvier #37
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… successfully being replaced the old value is still searchable
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Really nice work, you hit most of the learning goals here. Very elegantly written code. The only issue is with height
, as it's not quite working. Check out my comments and let me know any questions you have.
# Time Complexity: log(n) | ||
# Space Complexity: log(n) | ||
def add(key, value) |
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👍
# Time Complexity: log(n) | ||
# Space Complexity: log(n) | ||
def find(key) |
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👍
end | ||
end | ||
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def delete(key) |
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Just noting this isn't working.
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def remove(node) | ||
if node.left.nil? && node.right.nil? | ||
node = nil |
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This only changes the local variable (node) reference.
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
def inorder |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
def preorder |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
def postorder |
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👍
# Time Complexity: O(n) | ||
# Space Complexity: O(log n) | ||
def height |
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This method isn't actually working. See my note below.
Think about this:
If the node is nil
then return 0
If the node is not 0, find the height of the left and the right.
Return the height of the bigger subtree and add one (for the current node).
lib/tree.rb
Outdated
end | ||
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height_helper(node.left, left + 1, right) | ||
height_helper(node.right, left, right + 1) |
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This is only returning the right side's height!
No description provided.