<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:
    unordered_set<TreeNode*>rootNodes;
    unordered_map<TreeNode*,vector<TreeNode*>>mp;
    int ans=0,nodeCnt=0;
    void addEdgeToGraph(TreeNode* node1, TreeNode* node2){
        mp[node1].push_back(node2);
        mp[node2].push_back(node1);
    }

    void traverseTree(TreeNode* root,TreeNode* parentNode){
        if(root==NULL) return;
        else{
            nodeCnt++;
            if(parentNode!=NULL){
                addEdgeToGraph(root,parentNode);
            }
            if(root->left==NULL && root->right==NULL){
                rootNodes.insert(root);
            }else{
                traverseTree(root->left,root);
                traverseTree(root->right,root);
            }
        }
    }
    void dfs(TreeNode *node, unordered_set<TreeNode*>&visSet, int dist,int curDist){
        
        if(curDist>dist) return;
        visSet.insert(node);

        for(auto x:mp[node]){
            if(visSet.find(x)==visSet.end()){
                if(curDist+1<=dist && rootNodes.find(x)!=rootNodes.end()){
                    ans++;
                }
                dfs(x,visSet,dist,curDist+1);
            }
        }
    }
   
    int countPairs(TreeNode* root, int distance) {
        if(root->left==NULL && root->right==NULL) return 0;

        //convert tree to graph
        traverseTree(root,NULL);

        for(auto x:rootNodes){
            // cout<<"dfs("<<x->val<<")"<<endl;
            unordered_set<TreeNode*>visSet;
            dfs(x,visSet,distance,0);
        }

        // for(auto x:mp){
        //     cout<<x.first->val<<" -> ";
        //     for(auto y: x.second){
        //         cout<<y->val<<" ";
        //     }cout<<endl;
        // }

        return ans/2;
    }
};
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