<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 T
{
    public:

    int c;
    unordered_map <TreeNode *, int> um;
};

class Solution 
{
    T * func (TreeNode * root, int d)
    {
        if (root == NULL)
        {
            T * t = new T (); 
            t -> c = 0;
            return t;
        }
        if (root -> left == NULL && root -> right == NULL)
        {
            T * t = new T (); 
            t -> c = 0;
            t -> um[root] = 0;
            return t;
        }
        T * t = new T (); 
        T * tl = func (root -> left, d);
        T * tr = func (root -> right, d);
        t -> c = tl -> c + tr -> c;
        for (auto it : tl -> um)
        {
            if (it.second + 1 > d)
            {
                continue;
            }
            t -> um[it.first] = it.second + 1;
        }
        for (auto it : tr -> um)
        {
            if (it.second + 1 > d)
            {
                continue;
            }
            t -> um[it.first] = it.second + 1;
        }
        int c1 = 0;
        for (auto it1 : tl -> um)
        {
            for (auto it2 : tr -> um)
            {
                if (it1 != it2)
                {
                    if (it1.second + 1 + it2.second + 1 <= d)
                    {
                        c1++;
                    }
                }
            }
        }
        cout << c1 << endl;
        t -> c += c1;

        return t;
    }

    public:

    int countPairs (TreeNode * root, int distance) 
    {
        T * t = func (root, distance);

        return t -> c;
    }
};
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