Implement Binary Tree to Binary Search Tree Conversion #95
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--update : add test file for merge-two-sorted-arrays
BinarySearch test
2nd approach for product-of-elements
product-of-element.test.js
Re-implementation of Queue
Problems involving Queues
Add jumpSearch and Tests
* add solution to sort a stack problem * update README.md * add time compexity
BST unit test for delete()
Added tests for BST insertion
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@k88manish please go through the changes requested and make the changes where required.
| const tree = { | ||
| data: 10, | ||
| left: { data: 30, left: { data: 20 } }, | ||
| right: { data: 15, right: { data: 5 } }, | ||
| }; |
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Please follow the Node structure and adhere to it for all the things you do with BST.
Find the Node Implementation here: BST Node
| expect(tree).toEqual({ | ||
| data: 15, | ||
| left: { | ||
| data: 10, | ||
| left: { | ||
| data: 5, | ||
| }, | ||
| }, | ||
| right: { | ||
| data: 20, | ||
| right: { | ||
| data: 30, | ||
| }, | ||
| }, | ||
| }); | ||
| }); |
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Use the tree traversal method to verify the result
| function storeInorder(root, inorder) { | ||
| // Base case | ||
| if (!root) return; | ||
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| // First store the left subtree | ||
| storeInorder(root.left, inorder); | ||
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| // Append root's data | ||
| inorder.push(root.data); | ||
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| // Store right subtree | ||
| storeInorder(root.right, inorder); | ||
| } |
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Try to return an array instead of accepting an array as an argument. For help, you can go through the various traversal algorithms and problems on BST from the README
| storeInorder(root.right, inorder); | ||
| } | ||
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| // Helper function that copies of sorted array |
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Need more detailed information here
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@TheSTL please look into this. |
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@ashokdey I have made some changes as per your review. Please review it whenever possible. |
- Fixed ESlint errors - throw error for invalid string
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Closing this PR because of messed up Author info. Will open a new one for the same |
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