1457. Pseudo-Palindromic Paths in a Binary Tree

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

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:

C++

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;
    }

};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++