Just a heads up. Here is how I would recommend solving the problem using my reroot template:

default = [0] * n
   
def combine(nodeDP, neiDP, node, eind):
    return max(nodeDP, neiDP)
   
def finalize(nodeDP, node, eind):
    deg = len(graph[node]) - (eind >= 0)
    if deg >= 3:
        return nodeDP + 1
    elif deg == 2:
        return 1
    else:
        return 0

rootDP, forwardDP, reverseDP = rerooter(graph, default, combine, finalize)
print(max(rootDP))

Note that the variable deg inside finalize is the degree of the node in its subtree, where eind is the pointer to the parent of the node (which is -1 if the node is the root of the entire tree). In the end rootDP[node] contains the length of the longest centipede path ending at node. So max(rooDP) contains the length of the longest centipede path.

degree=2 can form a chain of length 1. I tunnel visioned on handling degree 3 and >3 only. didn't notice n>=3 gaurantees one degree 2 node.

I was able to do it without casework. Hint: