The Pseudo-Palindromic Paths in a Binary Tree problem counts root-to-leaf paths where the values along the path can form a palindrome (at most one value has an odd frequency). This coding problem uses DFS with a bitmask to track character frequencies. The BFS, DFS, bit manipulation, binary tree, and tree interview pattern is demonstrated.
Microsoft, Amazon, and Adobe ask this because it combines DFS tree traversal with the palindrome frequency check — and the elegant bitmask approach makes the frequency check O(1). The XOR toggle for each digit value is the key bit manipulation insight.
DFS with XOR bitmask. Maintain a bitmask where bit d is set if digit d has appeared an odd number of times. At each node: XOR the current node's value bit: mask ^= (1 << node.val). At a leaf: check mask & (mask-1) == 0 (at most one bit set = palindrome possible). Recurse on children. Backtrack: XOR again to restore.
Tree: 2→(3→(3,1),1→(1,None)). Path 2→3→3: mask=0 (all even). 0 & (-1)==0 ✓. Count. Path 2→3→1: mask after: 2^3^1 (bits 1,2,3 set? no: 2=bit2, 3=bit3, 1=bit1. mask=0b1110. popcount=3: not at most 1 bit → not pseudo-palindromic. Path 2→1→1: mask=bit2 only (2 appears once, two 1s cancel). 1 bit → ✓ Count. Total = 2.
Pseudo-Palindromic Paths elegantly combines two patterns: DFS traversal and bitmask frequency tracking. The XOR approach automatically tracks odd/even frequencies — XOR toggles each bit. At a leaf, check mask & (mask-1) == 0 to verify at most one odd-frequency digit. Practice similar "DFS with state tracking" problems: "count paths with XOR equal to k," "pseudo-random paths," "parity paths in trees."
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Sum of Nodes with Even-Valued Grandparent | Medium | Solve | |
| Find Bottom Left Tree Value | Medium | Solve | |
| Find the Level of Tree with Minimum Sum | Medium | Solve | |
| Reverse Odd Levels of Binary Tree | Medium | Solve | |
| Count Good Nodes in Binary Tree | Medium | Solve |