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

 

Solutions

Solution: Depth-First Search

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
const pseudoPalindromicPaths = function (root) {
  const getPalindromicPaths = (node = root, hash = new Set()) => {
    if (!node) return 0;
    const { val } = node;
    const isPalindromic = hash.has(val);

    isPalindromic ? hash.delete(val) : hash.add(val);
    const left = getPalindromicPaths(node.left, hash);
    const right = getPalindromicPaths(node.right, hash);
    const isPalindromicPath = hash.size < 2;

    isPalindromic ? hash.add(val) : hash.delete(val);
    if (!node.left && !node.right) return isPalindromicPath ? 1 : 0;
    return left + right;
  };

  return getPalindromicPaths();
};