Sabrina #32
Sabrina #32
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| # Time Complexity: O(log n) | ||
| # Space Complexity: O(n) | ||
| def add(key, value) |
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Since the number of recursive calls never go above the depth of the tree, it's O(log n) for space complexity due to the recursive call stack.
This of course assumes the tree is balanced.
| # Time Complexity: O(log n) | ||
| # Space Complexity: O(n) | ||
| def find(key) |
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Since the number of recursive calls never go above the depth of the tree, it's O(log n) for space complexity due to the recursive call stack.
This of course assumes the tree is balanced.
| # Time Complexity: | ||
| # Space Complexity: | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(2n) or O(n) | ||
| def height |
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👍 , but again the space complexity is O(log n) for a balanced tree and O(n) for an unbalanced one.
| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def postorder |
| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def preorder |
| # Time Complexity: O(n) | ||
| # Space Complexity: O(n) | ||
| def inorder |
| # else if the found node has 2 children | ||
| # - find the right child's leftmost leaf | ||
| right_child = current.right | ||
| leftmost_left = right_child.left |
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I think a better variable name here is leftmost_right = right_child.
Note I used right_child instead of right_child.left because you don't know if the right child has a left child as well.
| def bfs | ||
| raise NotImplementedError | ||
| end | ||
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| def delete(key) |
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I think this method is 90% there. When you have a node with 2 children to delete, you can swap the value of that node with it's smallest child on the right subtree, and then call delete again on the right subtree.
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