<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:
    int countPairs(TreeNode* root, int distance) {
        map<TreeNode*,vector<TreeNode*>>adj;
        queue<TreeNode*>q;
        set<TreeNode*>leaf;
        q.push(root);
        while(!q.empty()){
            TreeNode* top = q.front();
            q.pop();
            if(!top->left  && !top->right)  {
                leaf.insert(top);
                continue;
            }
            if(top->left) {
                adj[top].push_back(top->left);
                adj[top->left].push_back(top);
                q.push(top->left);
            }
            if(top->right)   {
                adj[top].push_back(top->right);
                adj[top->right].push_back(top);
                q.push(top->right);
            }
        }
        int ans=0;
        for(TreeNode*  l:leaf){
            int currDist=distance;
            queue<TreeNode*>ql;
            ql.push(l);
            // cout<<l->val<<"->"<<" ";
            unordered_map<TreeNode*,bool>vis;
            while(!ql.empty() && currDist>=0){
                currDist--;
                int s=ql.size();
                for(int i=0;i<s;i++){
                    TreeNode* top = ql.front();
                    // cout<<top->val<<", ";
                    if(leaf.find(top)!=leaf.end() && top!=l)  ans++;
                    ql.pop();
                    vis[top]=true;
                    for(TreeNode* x:adj[top]){
                        if(!vis.count(x))   ql.push(x);
                    }
                }
            }
            cout<<endl;
        }
        return ans/2;
        
    }
};

// 6-> 6, 1, 7, 4, 
// 5-> 5, 4, 7, 3, 
// 2-> 2, 3, 4, 7, 
110131