<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 makeGraph(TreeNode*root,unordered_map<TreeNode*,vector<TreeNode*>>& adj,unordered_map<TreeNode*,bool>& leaf,queue<TreeNode*>& q,TreeNode* parent){
        if(root==NULL){
            return;
        }
        if(parent!=NULL){
            adj[root].push_back(parent);
        }
        if(root->left){
            adj[root].push_back(root->left);
        }
        if(root->right){
            adj[root].push_back(root->right);
        }
        if(root->left==NULL && root->right==NULL){
            leaf[root]=true;
            q.push(root);
        }
        makeGraph(root->left,adj,leaf,q,root);
        makeGraph(root->right,adj,leaf,q,root);
    }
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*,bool>leaf;
        unordered_map<TreeNode*,vector<TreeNode*>>adj;
        queue<TreeNode*>q;
        makeGraph(root,adj,leaf,q,NULL);
        int count=0;
        while(!q.empty()){
            TreeNode* node=q.front();
            q.pop();
            cout<<node->val<<" LEAF"<<endl;
            queue<pair<TreeNode*,int>>qu;
            qu.push({node,0});
            unordered_map<TreeNode*,bool>visited;
            visited[node]=true;
            while(!qu.empty()){
                TreeNode* temp=qu.front().first;
                int step=qu.front().second;
                qu.pop();
                if(leaf[temp]==true && step<=distance && temp!=node){
                    count++;
                }
                for(auto it:adj[temp]){
                    if(!visited[it]){
                        if(step+1<=distance){
                        visited[it]=true;
                        qu.push({it,step+1});
                        }
                    }
                }
            }
        }
        return count/2;
    }
};
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