How do you solve Pseudo-Palindromic Paths in a Binary Tree on LeetCode?

Pseudo-Palindromic Paths in a Binary Tree (LeetCode #1457) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: A path can be pseudo-palindromic if at most one digit has an odd frequency.

What is the brute force solution for Pseudo-Palindromic Paths in a Binary Tree?

The brute force approach for Pseudo-Palindromic Paths in a Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all paths from the root to the leaf nodes and checking if each path can form a pseudo-palindrome. This method is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Pseudo-Palindromic Paths in a Binary Tree?

Explain your thought process clearly while coding. Consider edge cases, like single-node trees. Practice DFS and tree traversal techniques before the interview.

What are common mistakes when solving Pseudo-Palindromic Paths in a Binary Tree?

Not properly handling the leaf node check. Failing to reset the frequency count after backtracking.

#1457

Pseudo-Palindromic Paths in a Binary Tree

Medium

Bit ManipulationTreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchBacktrackingFrequency Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) while maintaining a count of the digit occurrences. This allows us to efficiently check for pseudo-palindromic paths without needing to store all paths explicitly.

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Algorithm

3 steps

1Step 1: Initialize a frequency array to count occurrences of digits from 1 to 9.
2Step 2: Perform a DFS from the root node, updating the frequency array as you traverse.
3Step 3: When reaching a leaf node, check the frequency array to see if at most one digit has an odd count.

solution.py22 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def pseudoPalindromicPaths(self, root: TreeNode) -> int:
10        def dfs(node, path_counts):
11            if not node:
12                return 0
13            path_counts[node.val] += 1
14            if not node.left and not node.right:
15                odd_count = sum(1 for count in path_counts if count % 2 == 1)
16                result = 1 if odd_count <= 1 else 0
17            else:
18                result = dfs(node.left, path_counts) + dfs(node.right, path_counts)
19            path_counts[node.val] -= 1
20            return result
21
22        return dfs(root, [0] * 10)

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the recursion stack in the worst case, where the tree is skewed.

1A path can be pseudo-palindromic if at most one digit has an odd frequency.
2Using a frequency array allows efficient counting without storing all paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.