^{5}.

The key observation in this problem is to imagine that we are building our tree step by step. Initially, we have only the root, which is also a leaf, so the sequence printed will be just 0. Now we add two children to the root, using edges with weights 0 and 1, and the depth-first search traverses them in some order. Depending on this order, the sequence we get will become either 0 1 or 1 0. We continue by repeatedly taking a leaf and adding two children to it, and the sequence always changes in the following manner: if the number corresponding to this leaf was *x*, then we insert *x*+1 into this sequence either directly to the left or directly to the right of it. And we can apply this operation to any number in the sequence, so we just need to check if the sequence we are given can be obtained starting from just the number 0 using this operation repeatedly.

Now let us look at the process in reverse, in other words we will repeatedly remove numbers that are adjacent to a number smaller by 1 until only 0 is left. What should we remove first? All numbers in this sequence that are adjacent to a number smaller by 1 are candidates. Now consider what happens when we remove number *x* that was adjacent to *x*-1: now *x*-1 becomes adjacent to some other number *y*. When *y*=*x*-2, the number *x*-1 become a candidate. When *y*=*x*, the number *y* becomes a candidate. In all other cases, no new candidates appear. This means that the new appearing candidates are never bigger than the number being removed.

Therefore we can safely remove any of the candidates with the highest value: we will never get candidates with even higher value in the future because of the previous argument, therefore those numbers themselves will never be useful to support removal of another number. So we can just keep removing one of the remaining highest values while it is possible, and we either reach a single 0, in which case we have found a solution, or we get stuck when all remaining highest values cannot be removed, in which case there is no solution.

Thanks for reading, and check back for more!

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