I even checked my calendar and it's not April -_-

Dear Tarun, I ask you not to prejudice and listen to me for a minute.

The problem created by RishabhKejariwal, he used "hair" in his statement. We have a grammar verification team. They decide the final statement. After they reviewed the statement, they chose to use "chocolate" instead of "hair." Unfortunately, they missed replacing all of the "hair"s to "chocolate"s.

If you have a link, publish it.

Solved with square-root decomposition and segment tree. Divide the euler tour of the tree into sqrt size blocks and sort each block. For each block maintain a segment tree where a node [l, r] stores the minimum depth of any node in this range.

For query of a node i, let the corresponding subtree be represented by subarray [l, r]. For partially intersecting blocks just do sqrt traversals and check if the given condition is satisfied. For fully intersecting blocks as these blocks where sorted previously, calculate two extreme indices j(let them be j1, j2) such that abs(A[i] — A[j]) >= K[i](which can be done with binary search). Now just query in the corresponding segment tree of that particular sqrt block in the range [0, j1] and [j2, sz(block) — 1].

Complexity — O(n√nlog√n)

It is evident that Euler tour will be necessary in order to access subtree of each node. Say we have the Euler tour with us in a list $$$E$$$. Now say for some node $$$u$$$ the subtree is from index $$$l$$$ to $$$r$$$ in $$$E$$$. The following steps will lead to finding the solution for $$$u$$$ :
1. Maintain a list of pairs $$$V$$$ of form $$$(A[E[idx]], Depth[E[idx]])$$$ for each $$$idx \,\epsilon\,[l,r]$$$.
2. Sort $$$V$$$. We now have a list of subtree elements of $$$u$$$ sorted on the basis of $$$A$$$ value.
3. Let $$$x$$$ be the rightmost position in $$$V$$$ where $$$A[x] \,\leq\,A[u]+k$$$ and $$$y$$$ be the leftmost position in $$$V$$$ where $$$A[y] \,\geq\,A[u]+k$$$. This can be done using binary search. So, $$$|A[u] - A[i]|\,\geq\, k \,\,\forall\,\, i\leq x$$$ and $$$|A[u] - A[i]|\,\geq\, k \,\,\forall\,\, i\geq y$$$.
4. We now just need to find minimum depth $$$\,\forall\,\, i\leq x$$$ and $$$\,\forall\,\, i\geq y$$$. This can be done by creating and prefix min and suffix min array.
Following the above steps for each node will lead to $$$O(N^{2}log(N))$$$ time complexity. So for optimisation apply $$$SQRT$$$ decomposition by dividing $$$A$$$ into $$$\sqrt{N}$$$ sized blocks, and maintaining data structures mentioned in $$$1 - 4$$$ for each block.
Now for each subtree from $$$[l,r]$$$ processing can be done in $$$(|r - l|/\sqrt{N})log(\sqrt{N})$$$ and overall $$$O(N\sqrt{N}log(\sqrt{N}))$$$ complexity.
Clearly this is not the most optimal solution for this problem, but it's different from the editorial as you hope for.