Pseudo-Palindromic Paths in a Binary Tree

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:


Solution:

/**
 * 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 {
    int res = 0;
    public int pseudoPalindromicPaths (TreeNode root) {
        if (root == null) return res;
        int[] count = new int[10];
        pre(root, count);
        return res;
    }
    
    private void pre(TreeNode root, int[] count) {
        if (root == null) return;
        count[root.val] ++;
        if (root.left == null && root.right == null) {
            // System.out.println(Arrays.toString(count));
            boolean odd = false, isP = true;
            for (int i = 1; i < 10; i ++) {
                if (count[i] % 2 == 1) {
                    if (odd) {
                        isP = false;
                        break;
                    } else {
                        odd = true;
                    }
                }
            }
            if (isP) {
                res ++;
            }
        }
        pre(root.left, count);
        pre(root.right, count);
        count[root.val] --;
    }
}