Aimee 👾 - Fire 🔥 #2
Open
Conversation
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
CheezItMan
reviewed
Apr 7, 2021
Comment on lines
+16
to
20
| // Time Complexity: Worst case, O(n) (the BST would be a structure similar to a linked list, just a linear chain of values). The BST grows exponentionally with respect to each level, so the total number of levels with respect to the total number of nodes is the inverse of an exponential - this is logarithmic growth. On average, this would be O(log(n)) because we are iterating through each level. On each level we only check one node - performing a constant number of operations. | ||
| // Space Complexity: O(1). We do not create a copy of the BST, we point to a node, and add a node. | ||
|
|
||
| // based on the tests the keys are numeric and the values are names - therefore we should sort/look-up by key | ||
| add(key, value) { |
Comment on lines
+54
to
56
| // Time Complexity: Worst case, O(n) (the BST would be a structure similar to a linked list. The BST grows exponentionally with respect to each level, so the total number of levels with respect to the total number of nodes is the inverse of an exponential - this is logarithmic growth. On average, this would be O(log(n)) because we are iterating through each level. On each level we only check one node - performing a constant number of operations. | ||
| // Space Complexity: O(1). We do not create a copy of the BST, we point to a node, and return the value of the node whose input key matches the node's key, otherwise we return null. | ||
| find(key) { |
There was a problem hiding this comment.
👍 , correct time/space complexity given the unbalanced tree.
Comment on lines
+81
to
86
| // Time Complexity: O(n), one constant time operation for every node | ||
| // Space Complexity: O(n), the array will be of length n, and a recursive function will be called for every node, (n number of stacks) | ||
| // inorder: left, root, right | ||
|
|
||
|
|
||
| inorder() { |
Comment on lines
+118
to
121
| // Time Complexity: O(n), one constant time operation for every node | ||
| // Space Complexity: O(n), the array will be of length n, and a recursive function will be called for every node, (n number of stacks) | ||
| // preorder: root, left, right | ||
| preorder() { |
Comment on lines
+152
to
155
| // Time Complexity: O(n), one constant time operation for every node | ||
| // Space Complexity: O(n), the array will be of length n, and a recursive function will be called for every node, (n number of stacks) | ||
| // postorder: left, right, root | ||
| postorder() { |
Comment on lines
+186
to
+188
| // Time Complexity: O(n), every node has to be checked | ||
| // Space Complexity: O(n) - worst case, if the BST is one long chain (skewed tree) then there will be n number of calls down the stack. On average, if the BST is more balanced, this can be O(log(n)) because the number of calls down the call stack will be logarithmic with respect to the number of nodes. The number of recursive calls will be at most the height of the tree. | ||
| height(node=this.root) { |
Comment on lines
+203
to
205
| // Time Complexity: O(n), the while loop will run n number of times (depends on the number of nodes). | ||
| // Space Complexity: O(n), the result array will be of length n (it will contain information about all the nodes). The temporary (temp) array will never be longer than length n. | ||
| bfs() { |
There was a problem hiding this comment.
👍 The time complexity is O(n^2) because temp.shift() is an O(n) operation.
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.
This suggestion is invalid because no changes were made to the code.
Suggestions cannot be applied while the pull request is closed.
Suggestions cannot be applied while viewing a subset of changes.
Only one suggestion per line can be applied in a batch.
Add this suggestion to a batch that can be applied as a single commit.
Applying suggestions on deleted lines is not supported.
You must change the existing code in this line in order to create a valid suggestion.
Outdated suggestions cannot be applied.
This suggestion has been applied or marked resolved.
Suggestions cannot be applied from pending reviews.
Suggestions cannot be applied on multi-line comments.
Suggestions cannot be applied while the pull request is queued to merge.
Suggestion cannot be applied right now. Please check back later.
No description provided.