<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:
void dfs(TreeNode* root, vector<vector<TreeNode*>>& adj, vector<TreeNode*> path)
{
	if (!root) return;
	path.push_back(root);
	if (!root->left && !root->right)
	{
		adj.push_back(path);
		return;
	}
	dfs(root->left, adj, path);
	dfs(root->right, adj, path);
}

int countPairs(TreeNode* root, int distance)
{
	/*vector<TreeNode*> val;
	getLeafNode(val, root);
	int ans = 0;
	for (int i = 0; i < val.size() - 1; ++i)
	{
		for (int j = i + 1; j < val.size(); ++j)
		{
			TreeNode* ancestor = LCA(root, val[i], val[j]);
			int left = distanceBetweenNode(ancestor, val[i]);
			if (left >= distance) continue;
			int right = distanceBetweenNode(ancestor, val[j]);
			if (right >= distance) continue;
			if (left + right <= distance) ++ans;
		}
	}
	return ans;*/
	vector<vector<TreeNode*>> adj;
	vector<TreeNode*> path;
	dfs(root, adj, path);
	int ans = 0;
	for (int i = 0; i < adj.size() - 1; ++i)
	{
		for (int j = i + 1; j < adj.size(); ++j)
		{
			for (int k = 1; k < adj[i].size() && k < adj[j].size(); ++k)
			{
				if (adj[i][k] != adj[j][k])
				{
					int left = adj[i].size() - k;
					int right = adj[j].size() - k;
					if (left + right <= distance)
						++ans;
					break;
				}
			}
		}
	}
	return ans;
}
};
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