Tree Node Stones – Make Adjacent Difference Exactly 1

The title gives a constraint but not the full objective, so first ask whether each node already has stones, which values are allowed, whether we minimize edits or count assignments, and whether edges are directed. A common formulation roots the tree and uses dp[node][v]: the best answer in node's subtree when node receives value v. For each child, combine only child values v-1 and v+1, because every adjacent difference must be exactly one; add the local change cost if values are being modified. Compute in postorder and take the best root state. With V permitted labels, the direct cost is usually O(nV) because each transition has only two neighbors, and space is O(nV) or O(hV) if released bottom-up. Test leaves, impossible value ranges, negative labels if allowed, and overflow in accumulated cost.