Search

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:

  • The given binary tree will have between 1 and 10^5 nodes.

  • Node values are digits from 1 to 9.


class Solution {
    
    private int ans = 0;
    public int pseudoPalindromicPaths (TreeNode root) {
        int[] freq = new int[10];
        
        helper(root, freq);
        return ans;
    }
    
    private void helper(TreeNode root, int[] freq){
        if(root== null){
            return ;
        }
        
        freq[root.val]++;
        
        if(root.left == null && root.right == null){
            // this is a leaf node
            if(isPalindromicPermutation(freq)){
                ans++;
            }
        }
        helper(root.left, freq);
        helper(root.right, freq);
        
        freq[root.val]--;
        
    }
    
    private boolean isPalindromicPermutation(int[] freq){
        boolean oddFreqFound = false;
        for(int el: freq){
            if(el%2!=0){
                if(oddFreqFound){
                    return false;
                } else{
                    oddFreqFound = true;
                }
                
            }
        }
        
        return true;
    }
}


9 views0 comments

Recent Posts

See All

A string s is called good if there are no two different characters in s that have the same frequency. Given a string s, return the minimum number of characters you need to delete to make s good. The f

The numeric value of a lowercase character is defined as its position (1-indexed) in the alphabet, so the numeric value of a is 1, the numeric value of b is 2, the numeric value of c is 3, and so on.