LeetCode #1457 — MEDIUM

Pseudo-Palindromic Paths in a Binary Tree

Move from brute-force thinking to an efficient approach using bit manipulation strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation · Tree

Example 1

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

Example 2

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

Example 3

[9]

Step 02

Core Insight

What unlocks the optimal approach

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

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1457: Pseudo-Palindromic Paths in a Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int pseudoPalindromicPaths(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode root, int mask) {
        if (root == null) {
            return 0;
        }
        mask ^= 1 << root.val;
        if (root.left == null && root.right == null) {
            return (mask & (mask - 1)) == 0 ? 1 : 0;
        }
        return dfs(root.left, mask) + dfs(root.right, mask);
    }
}

// Accepted solution for LeetCode #1457: Pseudo-Palindromic Paths in a Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pseudoPalindromicPaths(root *TreeNode) int {
	var dfs func(*TreeNode, int) int
	dfs = func(root *TreeNode, mask int) int {
		if root == nil {
			return 0
		}
		mask ^= 1 << root.Val
		if root.Left == nil && root.Right == nil {
			if mask&(mask-1) == 0 {
				return 1
			}
			return 0
		}
		return dfs(root.Left, mask) + dfs(root.Right, mask)
	}
	return dfs(root, 0)
}

# Accepted solution for LeetCode #1457: Pseudo-Palindromic Paths in a Binary Tree
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pseudoPalindromicPaths(self, root: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode], mask: int):
            if root is None:
                return 0
            mask ^= 1 << root.val
            if root.left is None and root.right is None:
                return int((mask & (mask - 1)) == 0)
            return dfs(root.left, mask) + dfs(root.right, mask)

        return dfs(root, 0)

// Accepted solution for LeetCode #1457: Pseudo-Palindromic Paths in a Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn pseudo_palindromic_paths(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        fn dfs(root: Option<Rc<RefCell<TreeNode>>>, mask: i32) -> i32 {
            if let Some(node) = root {
                let mut mask = mask;
                let val = node.borrow().val;
                mask ^= 1 << val;

                if node.borrow().left.is_none() && node.borrow().right.is_none() {
                    return if (mask & (mask - 1)) == 0 { 1 } else { 0 };
                }

                return (dfs(node.borrow().left.clone(), mask)
                    + dfs(node.borrow().right.clone(), mask));
            }
            0
        }

        dfs(root, 0)
    }
}

// Accepted solution for LeetCode #1457: Pseudo-Palindromic Paths in a Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function pseudoPalindromicPaths(root: TreeNode | null): number {
    const dfs = (root: TreeNode | null, mask: number): number => {
        if (!root) {
            return 0;
        }
        mask ^= 1 << root.val;
        if (!root.left && !root.right) {
            return (mask & (mask - 1)) === 0 ? 1 : 0;
        }
        return dfs(root.left, mask) + dfs(root.right, mask);
    };
    return dfs(root, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.