<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 bfs(TreeNode*node,int distance,unordered_map<TreeNode*,bool>&leaves,
 unordered_set<TreeNode*>&seen, unordered_map<TreeNode*,vector<TreeNode*>>&mp,int&cnt){

seen.insert(node);
queue<pair<TreeNode*,int>>q;
q.push({node,0});
while(!q.empty()){
    TreeNode*Tnode=q.front().first;
    int dis=q.front().second;
    if(dis>distance)break;
    q.pop();
    if(Tnode!=node&&leaves.find(Tnode)!=leaves.end()&&dis<=distance)cnt++;
    for(auto it:mp[Tnode]){
              if(!seen.count(it)){
            seen.insert(it);
           
    q.push({it,dis+1});
    
    }
    
}
}


}
void fillmap(TreeNode*root, unordered_map<TreeNode*,vector<TreeNode*>>&mp){
    if(root==NULL)return;
    if(root->left){
        mp[root].push_back(root->left);
        mp[root->left].push_back(root);
    }
    if(root->right){
        mp[root].push_back(root->right);
         mp[root->right].push_back(root);
    }
    fillmap(root->left,mp);
    fillmap(root->right,mp);
}
void findleaf(TreeNode*root,  unordered_map<TreeNode*,bool>&leaves){
    if(root==NULL)return;
    if(root->left==NULL&&root->right==NULL)leaves[root]=true;
    findleaf(root->left,leaves);
    findleaf(root->right,leaves);
}
    int countPairs(TreeNode* root, int distance) {
       unordered_map<TreeNode*,bool>leaves;
       unordered_map<TreeNode*,vector<TreeNode*>>mp;
       fillmap(root,mp);
       findleaf(root,leaves);
       int cnt=0;
      for(auto it:leaves){
       unordered_set<TreeNode*>seen;
      
        bfs(it.first,distance,leaves,seen,mp,cnt);

      }return cnt/2;
       
    }
};
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