<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>

class Solution {
public:
    int cnt=0;
    array<int, 10> dfs(TreeNode* root, int d) {
        // x[k-1]=number of leafs with distance=k
        array<int, 10> x={0};
        // if (!root) return x;

        if (!root->left && !root->right) {
            x[0] = 1;// root is a leaf node
            return x;
        }

        array<int, 10> left, right;
        if (root->left && root->right) {
            left= dfs(root->left, d);
            right= dfs(root->right, d);
            for (int i=0; i <d-1; i++) 
                x[i+1] = left[i]+right[i];

            // Prefix sum avoids of nested-loop
            for (int i=1; i <= d-1; i++)
                left[i] += left[i-1];
            
            //root is the LCA, apply multiplication principle for counting
            for (int i = 0; i < d-1; i++) 
                cnt += left[i]* right[d-2-i];
            
            return x;
        } else if (!root->left && root->right) {
            right= dfs(root->right, d);
            for (int i=0; i <d-1; i++) 
                x[i+1] = right[i];
                return x;
        } else if (root->left && !root->right) {
            left = dfs(root->left, d);
            for (int i=0; i <d-1; i++) 
                x[i+1] = left[i];
                return x;
        }
        
        return x;
    }

    int countPairs(TreeNode* root, int distance) {
        dfs(root, distance);
        return cnt;
    }
};
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