<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:
   vector<TreeNode*>leafs;
unordered_map<TreeNode*,TreeNode*>mp;

int countPairs(TreeNode* root, int distance) 
{	
     fillmap(root);
     
     int count=0;
     unordered_set<TreeNode*>st;
	  
     for(int i=0;i<leafs.size();i++)
     {
     	dfs(leafs[i],st,distance,0,count);
     	st.clear();
	 }
	
	 return count/2;
}
void dfs(TreeNode* node,unordered_set<TreeNode*>&st,int &distance,int dis,int &count)
{
	if(node==NULL)
	return;
	
	if(dis > distance)
	return;
	
	if(st.find(node)!=st.end())
	return;
	
	if(dis <= distance and node!=NULL and node->left==NULL and node->right==NULL and dis!=0)  
    {
      count++;
    }
	
	
	st.insert(node);
	
	dfs(node->left,st,distance,dis+1,count);
	dfs(node->right,st,distance,dis+1,count);
	dfs(mp[node],st,distance,dis+1,count);
	
}
void fillmap(TreeNode* root)
{
   if(root==NULL)
   return;
   
   if(root!=NULL and root->left==NULL and root->right==NULL)  //it is a leaf
   {
      leafs.push_back(root);	
   }	
   
   if(root!=NULL and root->left!=NULL)
   {
   	 mp[root->left]=root;
   }
   if(root!=NULL and root->right!=NULL)
   {
   	 mp[root->right]=root;
   }
   fillmap(root->left);
   fillmap(root->right);
}
};
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