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spitsfire
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Jul 13, 2022
| else: | ||
| current = current.right | ||
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| # Time Complexity: o(1) |
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Time complexity for find will be O(log n) because you are still iterating through the BST, but only one half of it every time.
| result = self.inorder_helper(current.right, result) | ||
| return result | ||
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| def inorder(self): |
| result = self.preorder_helper(current.left, result) | ||
| result = self.preorder_helper(current.right, result) | ||
| return result | ||
| def preorder(self): |
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| # Time Complexity: | ||
| # Space Complexity: | ||
| def postorder(self): |
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| # Time Complexity: | ||
| # Space Complexity: | ||
| def height(self): |
| # Time Complexity: | ||
| # Space Complexity: | ||
| # Time Complexity: O(log(n)) | ||
| # Space Complexity: O(n) |
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since you aren't recursively calling this, the space complexity is O(1) because no new memory is dependent upon the size of the input
| current = current.right | ||
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| # Time Complexity: o(1) | ||
| # Space Complexity: O(n) |
There was a problem hiding this comment.
since you aren't recursively calling this, the space complexity is O(1) because no new memory is dependent upon the size of the input
| # Time Complexity: o(1) | ||
| # Space Complexity: O(n) | ||
| # search for a node in the tree | ||
| def find(self, key): |
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