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Nice work Nataliia. You hit the learning goals here. Nice work with BFS. Well done
| # Tree Exercise | ||
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| In this exercise you will implement, in Ruby, several Tree methods. | ||
| In this exercise you will implement, in Python, several Tree methods. |
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| # Time Complexity: O(log n) for balanced BST, O(n) for unbalanced | ||
| # Space Complexity: O(1) | ||
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| def add_helper(self, current, key, value): |
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👍 , however the space complexity is O(log n) if the tree is balanced.
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| # Time Complexity: O(log n) for balanced BST, O(n) for unbalanced | ||
| # Space Complexity: O(1) | ||
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| def find_helper(self, current, key): |
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👍 However due to the recursive stack, the space complexity is O(log n) for a balanced tree.
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| # Time Complexity: O(n), because we visit each node once | ||
| # Space Complexity: O(depth of the recursion) or O(h), where h is height | ||
| # of the tree -> O(n) in the worst case | ||
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| def inorder_helper(self, current, result): |
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| # Time Complexity: ^ | ||
| # Space Complexity: ^ | ||
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| def preorder_helper(self, current, result): | ||
| if current: | ||
| result.append(current.to_json()) | ||
| self.preorder_helper(current.left, result) | ||
| self.preorder_helper(current.right, result) |
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| # ----------- post-order ---------- | ||
| # Time Complexity: ^ | ||
| # Space Complexity: ^ | ||
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| def postorder_helper(self, current, result): |
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| # Time Complexity: O(log n) for balanced, O(n) for unbalanced | ||
| # Space Complexity: O(n) | ||
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| def height_helper(self, current): |
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| # # Time Complexity: O(n) | ||
| # # Space Complexity:O(n) | ||
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| def bfs(self): |
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