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… successfully being replaced the old value is still searchable
CheezItMan
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Feb 27, 2020
CheezItMan
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CheezItMan
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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.
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| # 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) |
| end | ||
| end | ||
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| def delete(key) |
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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.
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| # 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
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| 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!
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