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CheezItMan
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Well done Chelsea, you hit all the requirements. Nice work!
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| # Time Complexity: log n because it is binary to find where to add the node | ||
| # Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
| def add(key, value) |
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| # Time Complexity: log n due to the binary search | ||
| # Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
| def find(key) |
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| # Time Complexity: O(n) because it has to go through every node. | ||
| # Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
| def inorder |
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| # Time Complexity: O(n) because it has to go through every node. | ||
| # Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
| def preorder |
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The space complexity for these traversals is O(n) because we're building an array.
With recursion it would be O(n + n) = O(2n) = O(n) or average case O(n + log n) = O(n)
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| # Time Complexity: O(n) because it has to go through every node. | ||
| # Space Complexity: worst case O(n) because we hit every node, better average case like in a balanced tree it would be O(log n) | ||
| def postorder |
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👍 With the earlier note on the traversals.
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| # Time Complexity: O(n) because we hit eery node | ||
| # Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
| def height |
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