Is Pseudo-Palindromic Paths in a Binary Tree easy or hard?

The problem is rated Medium on LeetCode because it combines tree traversal with a palindrome frequency condition. The DFS part is straightforward, but recognizing the odd-frequency rule for palindromes is the key insight. Once that observation is made, the implementation becomes relatively direct.

Pseudo-Palindromic Paths in a Binary Tree Python/Java solution

Python solutions typically use DFS recursion with a list or bitmask to track digit frequencies along the path. Java implementations often use an integer array of size 10 and backtracking during recursion. Both approaches achieve O(n) time complexity and O(h) space for the recursion stack.

How to solve Pseudo-Palindromic Paths in a Binary Tree in O(n)?

Traverse the tree using depth-first search while maintaining a frequency structure (array or map) for digits on the current path. When a leaf node is reached, verify that no more than one digit has an odd count. If the condition holds, increment the valid path counter. Because each node is processed once, the overall complexity remains O(n).

What is the best approach for Pseudo-Palindromic Paths in a Binary Tree?

Depth-first search with a digit count tracker is the most common solution. While traversing from root to leaf, maintain counts of digits along the path and check if at most one digit has an odd frequency. This guarantees the digits can form a palindrome. The traversal visits each node once, resulting in O(n) time complexity.

Is Pseudo-Palindromic Paths in a Binary Tree asked at Google/Amazon/Meta?

Binary tree DFS problems with path constraints are common in interviews at companies like Amazon, Google, and Meta. This problem tests recursion, path tracking, and understanding of palindrome properties. Variants involving path sums or path validation appear frequently in technical interviews.

What data structure is used in Pseudo-Palindromic Paths in a Binary Tree?

The solution uses a binary tree traversal combined with a frequency tracking structure. Most implementations use either a fixed array of size 10 for digit counts or a hash map. The traversal itself is implemented using depth-first search with recursion or an explicit stack.

What is the time complexity of Pseudo-Palindromic Paths in a Binary Tree?

The optimal solution runs in O(n) time where n is the number of nodes in the tree. Each node is visited exactly once during DFS traversal. The space complexity is O(h) due to the recursion stack, where h is the height of the binary tree.

Description Hints 2 Solution Video 13:18

1457. Pseudo-Palindromic Paths in a Binary Tree

Medium68.3% acceptance2 companies

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.

Example 1:

Input: root = [2,3,1,3,1,null,1]
Output: 2 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 2:

Input: root = [2,1,1,1,3,null,null,null,null,null,1]
Output: 1 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic since [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 3:

Input: root = [9]
Output: 1

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
1 <= Node.val <= 9

Asked at 2 companies

Bcg Amazon

Topics

Hints open one at a time

Hint 1

Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity).

Hint 2

Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity).

13:18

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1# Definition for a binary tree node.
2# class TreeNode:
3#     def __init__(self, val=0, left=None, right=None):
4#         self.val = val
5#         self.left = left
6#         self.right = right
7class Solution:
8    def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:
9        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[2,3,1,3,1,null,1]

Input

[2,1,1,1,3,null,null,null,null,null,1]

Input

[9]

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