Distance Between Two Tree Nodes

You are given a rooted tree whose nodes are numbered from 1 through treeNodes. The arrays treeFrom and treeTo describe the undirected edges of the tree, and root identifies its root.
Given two node IDs source and target, return the number of edges on the unique path between them.
The serialized edge list supplies the tree structure without parent pointers.

Function
distanceBetweenNodes(treeNodes: int, treeFrom: int[], treeTo: int[], root: int, source: int, target: int) → int

Examples
Example 1
treeNodes = 7
treeFrom = [1,1,2,2,3,3]
treeTo = [2,3,4,5,6,7]
root = 1
source = 4
target = 7
return = 4
The unique path is 4 -> 2 -> 1 -> 3 -> 7, which contains 4 edges.
Example 2
treeNodes = 5
treeFrom = [1,1,3,3]
treeTo = [2,3,4,5]
root = 1
source = 3
target = 5
return = 1
Node 5 is adjacent to node 3, so their distance is 1.

The trick: the tree is undirected and the path is unique, so you don't need the root or an LCA at all. Build an adjacency list from treeFrom and treeTo, then run BFS from source and return the depth at which you reach target. One pass, O(n) time and space. The common pitfall is assuming treeFrom is always the parent of treeTo and building a directed graph. That breaks when source and target sit in different branches, like 4 and 7 in Example 1. Another slip is forgetting a visited set, so BFS walks back up to the parent and loops. Handle source equal to target by returning 0. The LCA approach with depths works too, but it needs parent tracking and more code. If the adjacency list build or BFS escapes you under pressure, StealthCoder is the hedge that keeps you moving during the live OA.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.