BST traversals #42
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BST traversals #42
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CheezItMan
reviewed
Dec 3, 2021
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| while current: | ||
| previous = current | ||
| if current.key > new_node.key: | ||
| current = current.left | ||
| else: | ||
| current = current.right | ||
| if previous.key > new_node.key: | ||
| previous.left = new_node | ||
| else: | ||
| previous.right = new_node |
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You are storing things that are less than the current node on the right side and elements greater than on the left side. This works, but you end up with an inverted tree.
That's why your postorder isn't working properly on the tests.
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| # Time Complexity: O(logn) | ||
| # Space Complexity: O(1) | ||
| def find(self, key): |
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| # Time Complexity: O(logn) | ||
| # Space Complexity: O(1) | ||
| def inorder(self): |
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👍 However the space and time complexities are O(n) due to the list you're building
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| # Time Complexity: O(logn) | ||
| # Space Complexity: O(1) | ||
| def preorder(self): |
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| # Time Complexity: O(logn) | ||
| # Space Complexity: O(1) | ||
| def postorder(self): |
| left_height = self.height_helper(node.left) | ||
| right_height = self.height_helper(node.right) | ||
| return max(left_height,right_height)+ 1 | ||
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| def height(self): |
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| def bfs_helper(self,node,bfs_list): |
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