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
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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;
}
};
