> For the complete documentation index, see [llms.txt](https://coding-9.gitbook.io/untitled/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://coding-9.gitbook.io/untitled/bit-masking/1457.-pseudo-palindromic-paths-in-a-binary-tree.md).

# 1457. Pseudo-Palindromic Paths in a Binary Tree

## Problem Statement

<br>

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.*

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/05/06/palindromic_paths_1.png)

<pre><code><strong>Input: root = [2,3,1,3,1,null,1]
</strong><strong>Output: 2 
</strong><strong>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).
</strong></code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/05/07/palindromic_paths_2.png)

<pre><code><strong>Input: root = [2,1,1,1,3,null,null,null,null,null,1]
</strong><strong>Output: 1 
</strong><strong>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).
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: root = [9]
</strong><strong>Output: 1
</strong></code></pre>

&#x20;

**Constraints:**

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

## Intuition

```
Approach:

Go till leaf nodes, 
We can maintain a mask represnting 0123456789 at each bit level
Now, we can just xor the values and get set bits

Now interesting concept,
n-(n-1) == 0
To check if only one set bit or not

11000
01111
------
11000

Not zero so multiple set bits

1000
0111
-----
0000

So, we can see here we require only one set bit for odd
```

### Links

<https://leetcode.com/problems/pseudo-palindromic-paths-in-a-binary-tree/description/?envType=daily-question&envId=2024-01-24>

### Video Links

<https://www.youtube.com/watch?v=P0bAPxMkjnk&t=32s&ab_channel=AryanMittal>

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int count = 0;
    void find_ans(TreeNode* node, int path) {
        if (!node) {
            return ;
        }

        path ^= (1 << node->val);

        if (!node->left and !node->right and (path & (path - 1)) == 0) {
            count ++;
        }

        find_ans(node->left, path);
        find_ans(node->right, path);
    }

    int pseudoPalindromicPaths(TreeNode* root) {
        find_ans(root, 0);

        return count;
    }

};


```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###