Leah - Space #27
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Leah - Space #27
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CheezItMan
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Sep 2, 2020
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| # Time Complexity: Olog(n) - only need to look at half the tree | ||
| # Space Complexity: O(1) | ||
| def add(key, value) |
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👍 Since you're doing this recursively, the space complexity is O(log n) if the tree is balanced and O(n) if it is unbalanced. Same for time complexities.
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| # Time Complexity: Olog(n) - only need to look at half the tree | ||
| # Space Complexity: O(1) | ||
| def find(key) |
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| # Time Complexity: O(n) - n is the number of nodes in the tree | ||
| # Space Complexity: O(n) - my thought is that since we're making a list of nodes to return, we'll have that many nodes stored in the list, but since the stack doesn't ever hold something that's also already in the list, there would never be more than n nodes stored at one time..? Let me know if that's on the right track. | ||
| # left, root, right (me btwn my children) | ||
| def inorder |
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Nice work doing this with loops and a stack. I guess you REALLY don't like recursion.
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| # Time Complexity: O(n) - n is the number of nodes in the tree | ||
| # Space Complexity: O(n) - same assumption as inorder | ||
| # root, left, right (me before my children) | ||
| def preorder |
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| # Time Complexity: O(n) - n is the number of nodes in the tree | ||
| # Space Complexity: O(n) - same assumption as inorder | ||
| # left, right, root (me after my children) | ||
| def postorder |
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| # Time Complexity: Olog(n) in the best case (e.g. if the tree is balanced), O(n) in the worst case (the tree is unbalanced and essentially a linked list) | ||
| # Space Complexity: O(1) | ||
| def height |
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👍 , Since you are doing this recursively the space complexity is O(log n) if the tree is balanced and O(n) otherwise.
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| # Time Complexity: O(n) - n is number of nodes | ||
| # Space Complexity: O(n) - queue will hold at most the entire tree | ||
| def bfs |
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