Palindrome Path Queries in a Tree

You are given an undirected tree with n nodes numbered from 0 to n - 1. The tree is rooted at node 0, and each node i has a lowercase English character labels[i].
You are also given a list of queries. For each query [u, v], consider all node characters encountered on the simple path from u to v, including both endpoints.
Return an integer array where the answer for each query is 1 if the characters on that path can be rearranged to form a palindrome, and 0 otherwise.

Function
palindromePathQueries(n: int, edges: int[][], labels: String, queries: int[][]) → int[]

Examples
Example 1
n = 5
edges = [[0,1],[0,2],[1,3],[1,4]]
labels = "ababa"
queries = [[3,4],[2,3],[2,4]]
return = [1,1,0]
For [3,4], the path characters are b,b,a, which can form bab. For [2,3], the characters are a,a,b,b, which can form a palindrome. For [2,4], the characters are a,a,b,a, with two odd character counts, so they cannot form a palindrome.
Example 2
n = 4
edges = [[0,1],[1,2],[1,3]]
labels = "abca"
queries = [[0,2],[2,3],[1,1]]
return = [0,0,1]
The first two paths contain three different characters, so neither can be rearranged into a palindrome. A path from a node to itself always contains one character.

Constraints
n == labels.length
edges.length == n - 1
edges forms a valid tree rooted at node 0.
labels contains only lowercase English letters.
queries[i].length == 2

The trick is parity. Give each letter one bit in a 26-bit mask. For every node, compute the XOR of labels from the root down to that node. The path from u to v has mask pref[u] ^ pref[v] ^ bit(label[lca]), because the LCA is cancelled out by the two prefixes and needs to be added back once. Then check popcount(mask) <= 1, or mask & (mask - 1) == 0. The common pitfall is forgetting the LCA bit, which breaks the answer when the LCA's letter matters, like the [2,4] case in example 1. Another pitfall is recomputing paths per query. Do a DFS or BFS from node 0 to build prefix masks and depths, then use binary lifting for LCA. Preprocessing is O(n log n) and each query is O(log n). If the LCA code goes sideways under pressure, StealthCoder is the hedge in the live OA.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.