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` the number of times any node will be added to the euler walk is equal to number of its children every tree + 1 for non leaf nodes and 2 times for leaves (for upper bound we can assume that number of times a node occurs in a euler walk is equal to Nc + 2 where NcI is the number of children of a node I) for each node we can safely assume that every child gives it a single contribution and it has 2 contribution from itself in its frequency in euler path

now every node will be child of exactly one node except root (neglect it for now ) and this will contribute one frequency of its parent so total N(number of nodes in tree ) will be added to the euler vector size, and for each node 2 will be added as per above analysis hence at most 2*N will be added to the euler vector so total euler vector size should not increase 3*N

hope it is correct tell me if I am wrong somewhere `

If by "euler walk" you're referring to the Eulerian path/cycle in a general graph, then the length of that is exactly the number of edges in the graph. If you're talking about the Euler tour of a tree, then the number of vertices in the tour is exactly $$$2|V|-1$$$.