# 1457. Pseudo-Palindromic Paths in a Binary Tree

## Description

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

## Solutions

Solution 1: DFS + Bit Manipulation

A path is a pseudo-palindromic path if and only if the number of nodes with odd occurrences in the path is $0$ or $1$.

Since the range of the binary tree node values is from $1$ to $9$, for each path from root to leaf, we can use a $10$-bit binary number $mask$ to represent the occurrence status of the node values in the current path. The $i$th bit of $mask$ is $1$ if the node value $i$ appears an odd number of times in the current path, and $0$ if it appears an even number of times. Therefore, a path is a pseudo-palindromic path if and only if $mask \&(mask - 1) = 0$, where $\&$ represents the bitwise AND operation.

Based on the above analysis, we can use the depth-first search method to calculate the number of paths. We define a function $dfs(root, mask)$, which represents the number of pseudo-palindromic paths starting from the current $root$ node and with the current state $mask$. The answer is $dfs(root, 0)$.

The execution logic of the function $dfs(root, mask)$ is as follows:

If $root$ is null, return $0$;

Otherwise, let $mask = mask \oplus 2^{root.val}$, where $\oplus$ represents the bitwise XOR operation.

If $root$ is a leaf node, return $1$ if $mask \&(mask - 1) = 0$, otherwise return $0$;

If $root$ is not a leaf node, return $dfs(root.left, mask) + dfs(root.right, mask)$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the 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;
}
if (root.left == null && root.right == null) {
return (mask & (mask - 1)) == 0 ? 1 : 0;
}
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int pseudoPalindromicPaths(TreeNode* root) {
function<int(TreeNode*, int)> dfs = [&](TreeNode* root, int mask) {
if (!root) {
return 0;
}
if (!root->left && !root->right) {
return (mask & (mask - 1)) == 0 ? 1 : 0;
}
};
return dfs(root, 0);
}
};

• # 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:
if root is None:
return 0
if root.left is None and root.right is None:

return dfs(root, 0)


• /**
* 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
}
if root.Left == nil && root.Right == nil {
return 1
}
return 0
}
}
return dfs(root, 0)
}

• /**
* 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;
}
if (!root.left && !root.right) {
return (mask & (mask - 1)) === 0 ? 1 : 0;
}
};
return dfs(root, 0);
}


• // 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::rc::Rc;
use std::cell::RefCell;

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 val = node.borrow().val;

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

return (
);
}
0
}

dfs(root, 0)
}
}