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Even and Odd

You are given a **simple, undirected, connected** graph with **n** nodes numbered from 1 to n and **m** edges and a starting node **r**. You have to simulate a modified DFS: For each node u, if u is at an odd distance from r, then visit u's neighbours in ascending order(in terms of their node number) else visit them in descending order. Distance is defined as the number of edges between the cur

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solution.cppC++17

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