Scissors laurel O #60
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CheezItMan
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Feb 1, 2022
binary_search_tree/tree.py
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def add(self, key, value = None): |
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👍 However the time complexity is O(n) for an unbalanced tree and O(log n) for a balanced tree.
Also the space complexity is the same due to the call stack (recursion).
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| # Time Complexity: O(log n) | ||
| # Space Complexity: O(1) | ||
| def find(self, key): |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def inorder(self): |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def preorder(self): |
binary_search_tree/tree.py
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(log n) | ||
| def height(self): |
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👍 Space complexity is right if the tree is balanced O(n) otherwise.
binary_search_tree/tree.py
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| # # Time Complexity: O(n) | ||
| # # Space Complexity: O(n) | ||
| def bfs(self): |
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👍 However since you're using an array as a queue the queue.pop(0) is an O(n) operation by itself making your BFS solution O(n^2). You can reduce this by using a deque or a linked list.
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