Number of Good Leaf Nodes Pairs

Given the root of a binary tree and an integer distance. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance.

Return the number of good leaf node pairs in the tree.

Example 1:

Input: root = [1,2,3,null,4], distance = 3
Output: 1
Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.

Example 2:

Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2
Explanation: The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.

Example 3:

Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1
Explanation: The only good pair is [2,5].

Example 4:

Input: root = [100], distance = 1
Output: 0

Example 5:

Input: root = [1,1,1], distance = 2
Output: 1

Constraints:

• The number of nodes in the tree is in the range [1, 2^10].

• Each node's value is between [1, 100].

• 1 <= distance <= 10

class Solution {
    private int res;

    public int countPairs(TreeNode root, int distance) {
        res = 0;
        helper(root, distance);
        return res;
    }

    private int[] helper(TreeNode node, int distance) {
        if (node == null)
            return new int[11];

        // arr[i]: the number of leaf nodes at a distance 'i' from current node
        int[] A = new int[11];
        // Leaf Node
        if (node.left == null && node.right == null) {
            // Setting A[1] = 1, for the parent node
            A[1] = 1;
            return A;
        }
        int[] left = helper(node.left, distance);
        int[] right = helper(node.right, distance);
        // find all nodes satisfying distance
        for (int i = 0; i <= 10; ++i) {
            for (int j = 0; j <= 10; ++j) {
                if (i + j <= distance)
                    res += (left[i] * right[j]);
            }
        }
        // increment all by 1, ignore the node distance larger than 10
        for (int i = 0; i <= 9; ++i) {
            A[i + 1] += left[i];
            A[i + 1] += right[i];
        }
        return A;
    }
}