The palindrome condition is the easy part to state: at most one character with an odd frequency.
Use a bitmask to represent the parity of letter counts along the path from the root to each node. For each node, compute the mask via DFS; then count pairs of nodes whose masks differ by at most one bit (i.e., XOR is 0 or a power of two). Use a hash map to count frequencies of masks and their one-bit variants to achieve O(n * 26) time.
Pro tip: Emphasize that the palindrome condition depends only on parity of character counts, not order. Mention that using a 26-bit integer mask is efficient and that the counting step can be optimized by iterating over 26 possible bit flips.
A string can be rearranged into a palindrome if and only if at most one character has an odd count. So for a path, we need the parity of each letter's count.
Assign each letter a bit position (0-25). For any path, the parity mask is the XOR of masks of edges along the path. For root-to-node paths, compute mask via DFS.
For any two nodes u and v, the mask of the path between them is mask[u] XOR mask[v]. So we need pairs where mask[u] XOR mask[v] has at most one bit set.
Use a hash map to count frequencies of each mask. For each node, add counts of masks that differ by 0 or 1 bit (i.e., same mask or mask XOR (1<<i) for i=0..25). Accumulate pairs.
Time O(n * 26) and space O(n). Handle n=1 (no pairs) and ensure unordered pairs are counted once (e.g., by iterating and adding before inserting current node's mask).
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