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A good first step might be to compute the answer for every leaf. Take a leaf $$$L$$$ and root the tree there. For every other vertex $$$V$$$, it will be removed before $$$L$$$ if and only if the whole subtree of $$$V$$$ has values smaller than the value in $$$L$$$. Let $$$s[V]$$$ be the size of the subtree of $$$V$$$. Then, we want $$$L$$$ to have the smallest value among $$$s[V]+1$$$ vertices (that subtree and vertex $$$L$$$ itself), so the probability of $$$L$$$ being the smallest is $$$\frac{1}{s[V]+1}$$$. Since this is the probability of $$$V$$$ being before $$$L$$$, you can sum this up to get the expected value of the place of $$$L$$$ — because with that probability the place of $$$L$$$ is increased by $$$1$$$.

This is $$$O(N)$$$ solution to find an answer for a single leaf. I don't know how hard it is to apply it for non-leaf vertices. Try it and let us know.