<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 {
    void makeGraph(TreeNode* root, TreeNode* prev, unordered_map<TreeNode*, vector<TreeNode*>> &g, 
                    unordered_set<TreeNode*> &st) {
        if(root == NULL)
            return;
        
        if(prev != NULL) {
            g[root].push_back(prev);
            g[prev].push_back(root);
        }
        if(root->left == NULL && root->right == NULL) {
            st.insert(root);
        }
        makeGraph(root->left, root, g, st);
        makeGraph(root->right, root, g, st);
    }
public:
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*, vector<TreeNode*>> g;
        unordered_set<TreeNode*> st;
        makeGraph(root, NULL, g, st);
        int count = 0;
        queue<pair<TreeNode*, int>> q;
        for(auto it : st) {
            unordered_set<TreeNode*> vis;
            q.push({it, 0});
            vis.insert(it);
            while(!q.empty()) {
                auto iter = q.front();
                q.pop();
                auto node = iter.first;
                int level = iter.second;
                if(level + 1 > distance) continue;
                for(auto child : g[node]) {
                    if(vis.find(child) == vis.end()) {
                        if(st.find(child) != st.end()) {
                            count++;
                        }
                        q.push({child, level + 1});
                        vis.insert(child);
                    }
                }
            }
        }
        return count / 2;
    }
};
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