My first instinct was to just walk up the path and count character frequencies, which works but is way too slow if you think about the constraints.
Use a bitmask to represent the parity of character counts along the root-to-node path, since a palindrome is possible iff at most one character has an odd count. For each node, count ancestors whose bitmask differs by at most one bit from the current node's bitmask using a hash map during a DFS. This yields an O(N * 26) solution.
Pro tip: Mention that the bitmask approach reduces the problem to counting pairs with Hamming distance ≤ 1, which can be done efficiently with a hash map. Also, clarify that the path must be a contiguous sub-path of the root-to-node path, so we need to consider all ancestors, not just the root.
Clarify that a palindrome can be formed if at most one character has an odd frequency. The tree is rooted at 0, and for each node, we consider all ancestors (including itself) that form a contiguous sub-path from some ancestor to the node.
Assign each character a bit position (0-25). For each node, compute a bitmask where the i-th bit is 1 if the character appears an odd number of times along the path from the root to that node.
During DFS, maintain a hash map from bitmask to the number of times it has been seen along the current path. For the current node, the number of valid ancestors is the sum of counts for the current bitmask and for each bitmask that differs by exactly one bit (flipping each of the 26 bits).
Add the current node's bitmask to the hash map, recurse into children, then remove it (backtrack) to ensure the map only reflects the current path.
Time complexity is O(N * 26) since each node checks 27 bitmasks. Space is O(N) for the hash map and recursion stack. Handle empty tree, single node, and nodes with no valid ancestors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.