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2 | Benq | 3580 |
3 | orzdevinwang | 3570 |
4 | cnnfls_csy | 3569 |
5 | Geothermal | 3568 |
6 | tourist | 3565 |
7 | maroonrk | 3530 |
8 | Radewoosh | 3520 |
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10 | jiangly | 3467 |
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1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
4 | nor | 159 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 150 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
Tree Diameter
Can someone explain the logic given below for this problem.
One easy to implement solution is using 2 Breadth First Searches (BFS). Start a BFS with a random node and store the last node encountered before search ends. This last node will definitely be one of the ends of the diameter (Why?). Now run a second BFS from this node and you will end on the other end of the diameter.
Thanks
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