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CheezItMan
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Well done Gloria. You hit the learning goals here. Nice work.
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| # Time Complexity: O(n) because looping through all the nodes | ||
| # Space Complexity: O(n) because you need to add a node for each key, value | ||
| def add(self, key, value = None): |
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👍 The time complexity is O(n log n ) if the tree is balanced and O(n) otherwise.
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| def create_dict(self, TreeNode): | ||
| return { "key": TreeNode.key, | ||
| "value": TreeNode.value } |
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Useful helper, maybe put this in the TreeNode class.
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| # Time Complexity: log(n) because we are examining only half of the nodes | ||
| # Space Complexity: O(1) because we are not allocating any memory | ||
| # self refers to object the class is making | ||
| def find(self, key): |
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👍 The time complexity is O(n log n ) if the tree is balanced and O(n) otherwise.
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| def inorder(self): |
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| # Time Complexity: 0(N) | ||
| # Space Complexity: Space Complexity: 0(N) depends on input size | ||
| def preorder(self): |
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| # Time Complexity: 0(N) | ||
| # Space Complexity: 0(N) depends on input size | ||
| def postorder(self): |
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| # Time Complexity: 0(1) constant time | ||
| # Space Complexity: 0(1) | ||
| def height(self): |
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👍 this is O(n) for time complexity and O(n) space complexity if the tree is unbalanced and O(log n) if the tree is balanced.
| # Before pop head get info of current node add dict to the final array | ||
| # Encue the children insert at the tail | ||
| # https://www.geeksforgeeks.org/level-order-tree-traversal/ | ||
| def bfs(self): |
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