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Nice work, you have the essential methods working well. You also made progress trying BST and delete, although they're not working. Well done
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| # Time Complexity: O(logn), at worst, the algorighm will be called O(logn) times to add the new node to the bottom of the tree. | ||
| # Space Complexity: O(1) the amount of data stored is constant, though the space required in memory for the stack trace will be O(logn) | ||
| def add(key, value) |
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| # Time Complexity: O(logn), at worst, the algorighm will be called O(logn) times to add the new node to the bottom of the tree. | ||
| # Space Complexity: O(1) the amount of data stored is constant, though the space required in memory for the stack trace will be O(logn) | ||
| def find(key) |
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| # Time Complexity: O(n), because the algorithm will have to visit every node on the tree to add them to the nodes_array list | ||
| # Space Complexity: O(n) because the size of the nodes_array list will vary linearly with the number of nodes in the tree. | ||
| def inorder |
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| # Time Complexity: O(n), because the algorithm will have to visit every node on the tree to add them to the nodes_array list | ||
| # Space Complexity: O(n) because the size of the nodes_array list will vary linearly with the number of nodes in the tree. | ||
| def preorder |
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| # Time Complexity: O(n), because the algorithm will have to visit every node on the tree to add them to the nodes_array list | ||
| # Space Complexity: O(n) because the size of the nodes_array list will vary linearly with the number of nodes in the tree. | ||
| def postorder |
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| # Time Complexity: O(n), because the algorithm will have to visit every node on the tree in order to determine the height of the tree | ||
| # Space Complexity: O(1) because the amount of data stored is constant, though the space required in memory for the stack trace will be O(n) | ||
| def height |
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| # Time Complexity: O(n), because the algorithm will have to visit every node on the tree to add them to the nodes_array list | ||
| # Space Complexity: O(n) because the size of the nodes_array list will vary linearly with the number of nodes in the tree. | ||
| def bfs |
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Unfortunately this is more like a preorder traversal, this is because you traverse the entire left subtree before you traverse the right subtree. Remember all nodes at level 2 should be next to each other in the array and level 3 etc.
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| # couldn't get this to work in time :( | ||
| def delete(key) |
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