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

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Solutions

Solution 1

Intuition

Code

class Solution {
private:
    int res = 0;
    void dfs(int mask, TreeNode* node) {
        if (!node) return;
        mask ^= 1 << node->val;
        dfs(mask, node->left);
        dfs(mask, node->right);
        if (!node->left && !node->right) {
            if (mask == 0 || (mask & (mask - 1)) == 0) res++;
        }
    }
public:
    int pseudoPalindromicPaths(TreeNode* root) {
        dfs(0, root);
        return res;
    }
};

Complexity

• Time: O(n)
• Space: O(h)