<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 make(TreeNode* root,unordered_map<TreeNode*,vector<TreeNode*>> &mp,
    set<TreeNode*> &leaf){
        if(root==NULL)return ;
        if(root->left==NULL && root->right==NULL){
            leaf.insert(root);
        }
        if(root->left != NULL){
            mp[root].push_back(root->left);
            make(root->left,mp,leaf);
            mp[root->left].push_back(root);
        }
        if(root->right != NULL){
            mp[root].push_back(root->right);
            make(root->right,mp,leaf);
            mp[root->right].push_back(root);
        }
    }
    
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*,vector<TreeNode*>> mp;
        set<TreeNode*> leaf;
        make(root,mp,leaf);
        // cout<<leaf.size()<<endl;
        int ans=0;
        
        for(int i=0;i<leaf.size();){
            TreeNode* node=*(leaf.begin());
            // cout<<leaf.size()<<endl;
            leaf.erase(node);
            queue<pair<TreeNode*,int>> q;
            unordered_map<TreeNode*,int> visited;
            visited[node]=1;
            q.push({node,0});
            while(!q.empty()){
                auto it=q.front();
                q.pop();
                TreeNode* cn=it.first;
                visited[cn]=1;
                int di=it.second;
                if(di>distance)break;
                if(leaf.find(cn) != leaf.end()){
                    cout<<cn->val<<endl;
                    ans++;
                }
                for(auto &it : mp[cn]){
                    if(visited[it]==0){
                        q.push({it,di+1});
                    }
                }
            }
        }
        return ans;
    }
};
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