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Medium Tree DFS

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2020-05-23

1457. Pseudo-Palindromic Paths in a Binary Tree

Similar Question: LeetCode Question 266

Question:

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.

Solution:

Using the similar idea as LeetCode Question 266, when finding the path to leaves by using DFS, decide if the permutaion of the path is palindrom.

/**
 * 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 counter = 0;
    public int pseudoPalindromicPaths (TreeNode root) {
        dfs(root, new int[10]);
        return counter;
    }
    
    private void dfs(TreeNode node, int[] hash) {
        if(node == null) return;
        hash[node.val]++;
        if(node.left == null && node.right == null) {
            if(isPalindrome(hash)) counter++;
            hash[node.val]--;
            return;
        }
        dfs(node.left, hash);
        dfs(node.right, hash);
        hash[node.val]--;
    }
    
    // Similar to Leetcode 266
    private boolean isPalindrome(int[] hash) {
        int oddcounter = 0;
        for(int x : hash) {
            if (x % 2 == 1){
                oddcounter++;
            }
            if (oddcounter > 1) return false;
        }
        return true;
    }
}