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).