<p>You are given the <code>root</code> of a binary tree and an integer <code>distance</code>. A pair of two different <strong>leaf</strong> nodes of a binary tree is said to be good if the length of <strong>the shortest path</strong> between them is less than or equal to <code>distance</code>.</p>

<p>Return <em>the number of good leaf node pairs</em> in the tree.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/09/e1.jpg" style="width: 250px; height: 250px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4], distance = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/09/e2.jpg" style="width: 250px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6,7], distance = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the <code>tree</code> is in the range <code>[1, 2<sup>10</sup>].</code></li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= distance &lt;= 10</code></li>
</ul>

constexpr size_t max_distance = 11;

class Solution {
    int ans;
    size_t dist;
    vector<int> dfs(TreeNode *node) {
        if (!node) return {};
        if (!node->left and !node->right) return {1};
        vector<int> children_dist(max_distance);
        vector<int> left(dfs(node->left)), right(dfs(node->right));
        if (left.size() < max_distance) left.resize(max_distance, 0);
        if (right.size() < max_distance) right.resize(max_distance, 0);
        for (size_t i = max_distance - 1; i > 0; i--) {
            left[i] = left[i - 1];
            right[i] = right[i - 1];
        }
        two_sum(left, right);
        for (size_t i = 1; i < max_distance; i++) children_dist[i] = left[i] + right[i];
        return std::move(children_dist);
    }
    void two_sum(const vector<int> &left, const vector<int> &right) {
        // to be optimized
        for (size_t i = 1; i < max_distance; i++) {
            for (size_t j = 1; j < max_distance; j++) {
                if (i + j > dist) break;
                ans += left[i] * right[j];
            }
        }
    }
public:
    int countPairs(TreeNode *root, int distance) {
        dist = static_cast<size_t>(distance);
        ans = 0;
        dfs(root);
        return ans;
    }
};
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