Richelle ~~ water ~~ binary search tree #5
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CheezItMan
reviewed
Apr 15, 2021
| // Time Complexity: ? | ||
| // Space Complexity: ? | ||
| // Time Complexity: O(log(n) assuming the tree is balanced, because it would eliminate half the tree in each loop. If not balanced, then it would be O(n), acting more like a linkedList. | ||
| // Space Complexity: O(1) | ||
| add(key, value) { |
| // Time Complexity: ? | ||
| // Space Complexity: ? | ||
| // Time Complexity: O(log(n)) - if a balanced tree, otherwise, O(n) | ||
| // Space Complexity: O(1) | ||
| find(key) { |
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| // Time Complexity: O(n) | ||
| // Space Complexity: O(n) | ||
| // preorder: current, left, right (parent nodes get pushed to list first) | ||
| preorderHelper(current, array) { |
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| // Time Complexity: O(n) - touching every node | ||
| // Space Complexity: O(n) - array is length n, and n number of recursive calls on the stack | ||
| // inorder: left, current, right (left most nodes get pushed to list first, right last since it's largest) | ||
| inorderHelper(current, array) { |
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| // Time Complexity: O(n) - touches every node | ||
| // Space Complexity: O(log(n)) - for the call stack? | ||
| height(current=this.root, count=0) { |
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🤔 I don't see where this is evaluating which is larger the left or right subtree. It seems to just return the count if the left subtree exists.
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Sorry I think I must have cloned the base repo initially instead of forking it, well here is my submission now. Thanks! -Richelle