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CheezItMan
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Well done Delia, you hit the learning goals here. Nice work!
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| # Time Complexity: O(log n) *if balanced | ||
| # Space Complexity: O(1) | ||
| def add(self, key, value = None): |
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👍 Good note about being balanced.
nice iterative solution
| else: | ||
| parent.right = TreeNode(key, value) | ||
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| def add_helper(self, curr_node, key, value): |
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| if self.root == None: | ||
| self.root = TreeNode(key, value) | ||
| else: | ||
| # use helper method to add parameter | ||
| self.add_helper(self.root, key, value) |
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Can be rewritten
Suggested change
| if self.root == None: | |
| self.root = TreeNode(key, value) | |
| else: | |
| # use helper method to add parameter | |
| self.add_helper(self.root, key, value) | |
| self.root = self.add_helper(self.root, key, value) |
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| # Time Complexity: O(log n) *if balanced | ||
| # 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 postorder(self): |
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def height(self): |
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👍 The space complexity is O(h) where h is the height of the tree, but you're right if it's unbalanced h could be the number of nodes.
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| # # Time Complexity: O(n) | ||
| # # Space Complexity: O(n) | ||
| def bfs(self): |
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