I recently thought of this interesting Tree problem:

Given a tree with $N$ nodes, let $val_i$ = the "value" for each node. Process $Q$ queries of two types: the first consisting of a single node $i$. For this type of query, $val_j += dist(i, j)$ for every other node $j$, where $dist(i, j)$ = the distance between nodes i and j in the tree. The second type of query just asks for $val_j$ of some node $j$.

It seems to me that would be a well-known problem, but I have not been able to find any reference or mention of this problem online. Currently, my best approach is to simply brute-force; when given a query $i$, just run a DFS from node $i$ and update all the other nodes manually in $O(N)$. Of course, I am hoping for a more efficient algorithm.

Can anyone suggest one such algorithm/strategy?

Some more ideas: Lazy Propagation (to avoid having to update $O(N)$ values immediately), Heavy-Light Decomposition/Centroid Decomposition (not really any ideas for how these will be applied)...

  • $\begingroup$ Is the tree balanced? Do you have to keep it balanced after each query as well? or do you need only to update the "values" (without reordering the tree)? $\endgroup$
    – nir shahar
    Dec 21, 2021 at 18:40
  • $\begingroup$ I'm sorry, but I don't quite understand. The tree is given at the beginning, before queries are to be answered. The tree's structure itself doesn't change. $\endgroup$ Dec 21, 2021 at 18:41
  • $\begingroup$ Sorry, I forgot to answer your first question. The tree is not necessarily balanced (if balanced means that the sizes of the left and right subtrees of any node differ by one or less). Do you think it needs to be? I still can't think of an algorithm for balanced trees. $\endgroup$ Dec 21, 2021 at 18:46
  • $\begingroup$ Since you have to modify $\Omega(N)$ values in the tree for the first query, isn't the DFS already your best bet? $\endgroup$
    – Nathaniel
    Dec 21, 2021 at 18:46
  • $\begingroup$ @Nathaniel it sounds like the OP has to do $Q$ such queries, and is hoping for complexity less than $\Omega(NQ)$ overall $\endgroup$
    – nir shahar
    Dec 21, 2021 at 18:48

1 Answer 1


There are two ways this problem can be solved. I really like these solutions.

  1. Square Root Decomposition

Keep a set of at most O(sqrtN) "active nodes". When you have a query of type 2, just loop through all these active nodes and add the distance between each one to the queried node. When you have a query of type 1, add the new node to the set if the size does not exceed sqrtN. When the size of this set exceeds sqrtN, we need to basically somehow empty the set and update all nodes with their total distance to all active nodes. But this can be easily done in O(N), not O(NsqrtN). The idea is that let's say the root of the tree is node 0, and first find the total distance sum of all active nodes to the root in O(N). Then, consider a node n and its child c, and let ans[n] = the total distance sum of all active nodes to node n. When we move from node n to its child c, we move one closer to every active node in the subtree of c, while moving one further from every other active node in the rest of the tree. So dp[c] = dp[n] - (sub[c]) + (total number of active nodes - sub[c]), where sub[c] is the number of active nodes in the subtree of c (and all sub[..] can be precomputed in O(N), with the same approach one would use to find subtree sizes). So this transition is O(1), meaning that we need O(N) time to simultaneously apply all updates (note that this works for any number of active nodes). In the end, the total time complexity is O(QsqrtN) (since each query of type 2 is O(sqrtN) in the worst case, and queries of type 1 are O(Q/sqrtN * N) = O(QsqrtN) amortized). BUT: Overall, this is quite slow for me, taking 5 or 6 seconds to process 200k queries over a tree with 100k nodes.

One example of this sort of reasoning: https://cses.fi/problemset/task/1133 https://usaco.guide/problems/cses-1133-tree-distances-ii/solution

  1. Heavy Light Decomposition

Use the same recurrence as above with the same definitions: dp[c] = dp[n] + sz - 2 * sub[c], where sz = total number of active nodes in the whole tree. Now, what happens if we substitute the recurrence in for dp[n]? Then we get dp[c] = (dp[parent of n] + N - 2 * sub[n]) + N - 2 * sub[c]. Reapplying this recurrence until we reach the root (say, node 0) gets us

dp[c] = dp[0] + (depth of node c) * sz - 2 * (for every node i on the path from c to the root, except for the root itself) sub[i].


It's clear that dp[0] is easy to maintain (when we get a type 1 query with a new node, just add its distance from the root, which just equals its depth, to dp[0]). sz is also trivial to maintain (it's just the total number of type 1 queries received up to a given point).

But how can we efficiently find 2 * (for every node i on the path from c to the root, except for the root itself) sub[i]?

Heavy Light Decomposition. Basically we need to support two types of queries:

  1. Update every value on a path to the root by 1
  2. Find the sum of all values on a path from the root (and then subtract out sub[root] since remember we don't count that in the expression)

And this can be done with Heavy-Light Decomposition incorporating a Lazy Segment Tree.

For the same bounds as above (200k queries on a tree of 100k nodes), I had a runtime of less than 300ms.


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