Vultr DocsLatest Content

Pseudo-Palindromic Paths in a Binary Tree

Updated on 10 July, 2025
Pseudo-Palindromic Paths in a Binary Tree header image

Problem Statement

In this problem, we are given a binary tree where each node's value is a digit between 1 and 9. Our task is to find the number of paths from the root node to any leaf that are pseudo-palindromic. A path is defined as pseudo-palindromic if at least one permutation of the node values in the path forms a palindrome. A palindrome is a sequence that reads the same forward and backward. As a result, we need to determine how many such paths exist in the given binary tree.

Examples

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

Approach and Intuition

  1. Understanding Pseudo-Palindromic Paths:
    A path is termed pseudo-palindromic if the digits in the path can be rearranged to form a palindrome. For a sequence to be rearranged into a palindrome:

    • All characters (or numbers in this case) must appear an even number of times, with the exception of at most one character, which can appear an odd number of times. This character, if present, would sit in the middle of the palindrome.

  2. Counting the Frequencies Using Bitwise Operations:
    To efficiently check if any permutation of the path's numbers can form a palindrome:

    • Use a bitwise technique to count the frequencies. Each bit in an integer can represent whether the count of a number (1 through 9) is odd or even.
    • Toggle the corresponding bit for a number using XOR each time it appears in the path. After processing all numbers in a path:
      • If the binary representation has more than one bit set (i.e., more than one number has an odd count), the permutation cannot form a palindrome.
      • Else, it can.

  3. Depth-First Search (DFS) Implementation:

    • Start a DFS from the root, carrying the current state of number frequencies (using the bitwise technique described).
    • At each node, toggle the bit corresponding to the node's value.
    • If a leaf node is reached, check the bit count of the frequencies. If it meets the palindrome criteria, increment a global counter.
    • Recursively process all paths from the root to leaves.

  4. Optimizations and Edge Cases:

    • If the tree has only one node, it’s automatically a pseudo-palindromic path.
    • Utilize early termination in DFS if it's already determined that a path cannot be turned into a palindrome path based on current bit counts.

This approach leverages bitwise operations for efficient frequency tracking and depth-first search for exploring all root-to-leaf paths, thus ensuring that all potential pseudo-palindromic paths are counted.

Solutions

  • Java
java 
class Solution {
    int sum = 0;
        
    public void traverse(TreeNode currentNode, int currentPath) {
        if (currentNode != null) {
            currentPath = currentPath ^ (1 << currentNode.val);
            if (currentNode.left == null && currentNode.right == null) {
                if ((currentPath & (currentPath - 1)) == 0) {
                    ++sum;
                }
            }
            traverse(currentNode.left, currentPath);
            traverse(currentNode.right, currentPath);
        }
    }
    
    public int findPseudoPalindromicPaths(TreeNode root) {
        traverse(root, 0);
        return sum;
    }
}

The problem addressed involves counting pseudo-palindromic paths in a binary tree using Java. A path in the tree is pseudo-palindromic if the sequence of node values it contains can be rearranged to form a palindrome.

The provided Java solution employs a depth-first search strategy for traversal and bitwise operations to efficiently track and evaluate the palindromic nature of paths:

  • Define a class Solution with a member variable sum to accumulate the number of pseudo-palindromic paths.
  • Implement a method traverse, which operates recursively. It takes two parameters: currentNode, representing the node currently under examination, and currentPath, which holds the binary presentation of the path from the root to the current node using bitwise XOR to toggle bits.
  • Use bitwise shift 1 << currentNode.val to manage path states. This shifts a binary 1 to the left by a position indicated by the node’s value, allowing toggling of the respective bits of currentPath.
  • Determine if currentNode is a leaf (i.e., it has no children) by checking if both currentNode.left and currentNode.right are null. At this point, determine if the path can form a palindrome:
    • A path can be rearranged into a palindrome if at most one bit in currentPath is set. This is checked using the condition (currentPath & (currentPath - 1)) == 0, which tests if currentPath is zero or a power of two.
  • If the conditions are met, increment sum.
  • Continue the recursion to both child nodes.
  • In findPseudoPalindromicPaths, initiate the recursive traversal starting from the root node with an empty path (0) and eventually return the count of valid paths.

The solution effectively manages state during the tree traversal without the need for additional storage structures, leaning on bitwise operations to track and verify path conditions efficiently. This approach provides a neat method to check for the pseudo-palindromic property while minimizing space complexity.

  • Python
python 
class Solution:
    def countPseudoPalindromicPaths(self, root: TreeNode) -> int:
        def dfs(node, bitmask):
            nonlocal total
            if node:
                # Toggle the current node's value in the bitmask
                bitmask ^= (1 << node.val)
                # Check if current node is leaf
                if not node.left and not node.right:
                    # Evaluate if path can represent a palindrome
                    if bitmask & (bitmask - 1) == 0:
                        total += 1
                else:
                    dfs(node.left, bitmask)
                    dfs(node.right, bitmask)
    
        total = 0
        dfs(root, 0)
        return total

The Python code implemented in the Solution class defines a method countPseudoPalindromicPaths to determine the number of pseudo-palindromic paths in a binary tree. Each path is considered pseudo-palindromic if the digits can be rearranged to form a palindrome.

  • Use depth-first search (DFS) to traverse each path in the binary tree.
  • Maintain a bitmask to record the occurrence of each digit using bitwise operations.
  • For each node, toggle the corresponding bit in the bitmask using XOR.
  • On reaching a leaf node, check if the current path could form a palindrome. This is determined by checking if there is at most one bit set in the bitmask.
  • Accumulate the count of pseudo-palindromic paths that meet the criteria.

The method returns the total count of such paths starting from the root node. This approach utilizes bitwise operations for efficient tracking and checking of potential pseudo-palindromic sequences as the tree is traversed.

Comments

No comments yet.