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by Freire_Herval
1149 days ago
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There's an error in my calculations. Apologies. The comparison happens at every level and NOT every node so, it's O(N) for traversal then when a layer is completed a comparison occurs at a cost of O(N^2). The amount of times this happens is equal to the height of the tree which can be simplified to logN. So total big O is O(N) + O(N^2 * log2(N)). This simplifies to O(N^2 * log2(N)) which, while better than O(N^3), is still really bad. |
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