Pseudo-Palindromic Paths in a Binary Tree - Problem
Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.
Return the number of pseudo-palindromic paths going from the root node to leaf nodes.
Note: A palindrome reads the same forward and backward. For example, 121 is a palindrome while 123 is not.
Input & Output
Example 1 — Basic Tree
$
Input:
root = [2,3,1,3,1,null,1]
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Output:
2
💡 Note:
Path [2,3,3] has digits 2(×1), 3(×2) - only one odd count, can form palindrome '323'. Path [2,1,1] has digits 2(×1), 1(×2) - only one odd count, can form palindrome '121'. Path [2,3,1] has all odd counts - cannot form palindrome.
Example 2 — Single Path
$
Input:
root = [2,1,1,1,3,null,null,null,null,null,1]
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Output:
1
💡 Note:
Only one root-to-leaf path exists: [2,1,1,1]. Digit frequencies: 2(×1), 1(×3). Only one digit has odd count, so palindrome '11211' is possible.
Example 3 — No Palindromes
$
Input:
root = [9]
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Output:
1
💡 Note:
Single node path [9] has only one digit with count 1 (odd). Single digits are always palindromes.
Constraints
- The number of nodes in the tree is in the range [1, 105]
- 1 ≤ Node.val ≤ 9
Visualization
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Understanding the Visualization
1
Input Tree
Binary tree with digit values 1-9
2
Path Analysis
Check each root-to-leaf path for palindrome potential
3
Count Result
Return number of valid pseudo-palindromic paths
Key Takeaway
🎯 Key Insight: Use bit manipulation to efficiently track odd/even digit frequencies during DFS traversal
💡
Explanation
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