<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>

//-------------------This is not most optimal.. will do it.. but this works too------

/**
 * 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:
    vector<TreeNode*> leaves;
    unordered_map<TreeNode*,vector<TreeNode*>> adj;
    int count=0;

    void helperCountPairBfs(TreeNode* leaf, int distance){
        unordered_map<TreeNode*, int> visited;
        queue<TreeNode*> q;
        q.push(leaf);
        visited[leaf] = 1;
        int level = 0;

        while(!q.empty()){
            int n = q.size();
            while(n--){
                auto temp = q.front();
                q.pop();
                
                if(!temp->left && !temp->right && temp!=leaf){
                    // cout<<leaf->val<<" "<<temp->val<<endl;
                    count++;
                }
                for(auto i : adj[temp]){
                    if(!visited[i]){
                        visited[i]=1;
                        q.push(i);
                    }
                }

            }
            level++;
            if(level>distance) break;
        }
    }

    void makeGraph(TreeNode* root, TreeNode* parent){
        if(!root) return ;

        if(parent){
            adj[root].push_back(parent);
            adj[parent].push_back(root);
        }

        makeGraph(root->left, root);
        makeGraph(root->right, root);

        if(!root->left && !root->right) leaves.push_back(root);
    }

    int countPairs(TreeNode* root, int distance) {
        if(!root || (!root->left && !root->right)) return 0;

        makeGraph(root, 0); // parent=0 as root.. given node value starts from 1
        
        // for (auto temp : adj) {
        //     TreeNode* node = temp.first;
        //     cout << "Node " << node->val << " is connected to: ";
        //     for (auto i : adj[node]) {
        //         cout << i->val << " ";
        //     }
        //     cout << endl;
        // }

        for(auto i : leaves){
            helperCountPairBfs(i, distance);
        }

        return count/2;
    }
};
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