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CheezItMan
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Apr 27, 2021
CheezItMan
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CheezItMan
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Nice work Roshni, you hit the learning goals here. Well done.
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| # Time Complexity: O(log n) for balanced, O(n) for unbalanced | ||
| # Space Complexity: O(log n) for balanced, O(n) for unbalanced | ||
| def add(key, value = nil) |
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| # iterative add helper | ||
| # def add_helper(current, key, value) | ||
| # added_to_tree = false | ||
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| # until added_to_tree | ||
| # if key <= current.key | ||
| # next_node = current.left | ||
| # if next_node.nil? | ||
| # new_node = TreeNode.new(key, value) | ||
| # current.left = new_node | ||
| # added_to_tree = true | ||
| # else | ||
| # current = current.left | ||
| # end | ||
| # else | ||
| # next_node = current.right | ||
| # if next_node.nil? | ||
| # new_node = TreeNode.new(key, value) | ||
| # current.right = new_node | ||
| # added_to_tree = true | ||
| # else | ||
| # current = current.right | ||
| # end | ||
| # end | ||
| # end | ||
| # end |
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| # Time Complexity: O(log n) for balanced, O(n) for unbalanced | ||
| # Space Complexity: O(log n) for balanced, O(n) for unbalanced | ||
| def find(key) |
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| # Time Complexity: O(n) to visit all nodes | ||
| # Space Complexity: O(n) values array depends on how many nodes there are | ||
| def inorder |
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| # Time Complexity: O(n) to visit all nodes | ||
| # Space Complexity: O(n) values array depends on how many nodes there are | ||
| def preorder |
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| # Time Complexity: O(n) to visit all nodes | ||
| # Space Complexity: O(n) values array depends on how many nodes there are | ||
| def postorder |
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| # Time Complexity: O(n) to visit all the nodes | ||
| # Space Complexity: O(log n) if balanced, O(n) if unbalanced | ||
| def height |
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| # Time Complexity: O(n^2) due to use of .shift | ||
| # Space Complexity: O(n) | ||
| # Solution from class | ||
| def bfs |
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👍 This should work, and good insight with the .shift's affect on the time complexity.
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