Problem

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

Solution

/**
 * 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}
 */
var pseudoPalindromicPaths  = function(root) {
    return dfs(root, Array(10).fill(0), 0);
};

var dfs = function(node, map, numOfOdd) {
    if (!node) return 0;
    if (map[node.val] % 2) {
        numOfOdd -= 1;
    } else {
        numOfOdd += 1;
    }
    if (!node.left && !node.right) {
        return numOfOdd <= 1 ? 1 : 0;
    }
    map[node.val] += 1;
    var res = dfs(node.left, map, numOfOdd)
        + dfs(node.right, map, numOfOdd);
    map[node.val] -= 1;
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).