<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>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countPairs(TreeNode* root, int distance) {
        std::vector<int> bucket_counts_child1(distance, 0);
        return pairCount(root, distance, bucket_counts_child1);
    }

    int pairCount(TreeNode* root, int distance, std::vector<int>& bucket_counts) {
        int leftCount = 0, rightCount = 0;
        std::vector<int> bucket_counts_child1(distance, 0);
        std::vector<int> bucket_counts_child2(distance, 0);

        if(root->left)
            leftCount = pairCount(root->left, distance, bucket_counts_child1);
        if(root->right)
            rightCount = pairCount(root->right, distance, bucket_counts_child2);
        if(!root->left && !root->right) {
            bucket_counts[0] = 1;
            return 0;
        }
        
        int newPairsCount = 0;

        // update all distances from current root
        for(int i = 1; i < distance; i++) {
            bucket_counts[i] += bucket_counts_child1[i-1];
            bucket_counts[i] += bucket_counts_child2[i-1];
        }

        for(int i = 1; i < distance; i++) {
            for(int j = 1; j+i <= distance; j++) {
                newPairsCount += bucket_counts_child1[i-1] * bucket_counts_child2[j-1];
            }
        }
        
        return leftCount + rightCount + newPairsCount;
    }
};
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