<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:
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*, vector<TreeNode*>> adj;
        set<TreeNode*> st;

        makeParent(root, NULL, adj, st);

        int cnt = 0;
        for(auto leaf : st){
            queue<TreeNode*> q;
            set<TreeNode*> vis;
            vis.insert(leaf);
            q.push(leaf);

            int level = 0;
            while(!q.empty()){
                int sz = q.size();

                for(int i = 0;i < sz;i++){
                    auto x = q.front();
                    if(level <= distance && x != leaf && st.count(x)) cnt++;
                    q.pop();
                    for(auto nnode : adj[x]){
                        if(vis.find(nnode) == vis.end()){
                            q.push(nnode);
                            vis.insert(nnode);
                        }
                    }
                }
                level++;
                if(level > distance) break;
            }
        }
        return cnt/2;
    }
    void makeParent(TreeNode* root, TreeNode* prv,  unordered_map<TreeNode*, vector<TreeNode*>> &adj, set<TreeNode*> &st){
        if(!root) return;

        if(!root->left && !root->right) st.insert(root);

        if(prv != NULL){
            adj[root].push_back(prv);
            adj[prv].push_back(root);
        }

        makeParent(root->left, root, adj, st);
        makeParent(root->right, root, adj, st);
    }
};
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