<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 {
    unordered_map<TreeNode*, vector<TreeNode*>>mp;
    vector<TreeNode*>leaves;
    int ans = 0;
    
    void storeLeaves(TreeNode* root) {
        if (!root) return;
        storeLeaves(root->left);
        if (!root->left && !root->right)
            leaves.push_back(root);
        if (root->left) {
            mp[root].push_back(root->left);
            mp[root->left].push_back(root);
        }
        if (root->right) {
            mp[root].push_back(root->right);
            mp[root->right].push_back(root);
        }
        storeLeaves(root->right);
    }
    
    void bfs(TreeNode* v, int distance) {    
        queue<TreeNode*>q;
        q.push(v);
        int dist = 0;
        
        unordered_map<TreeNode*, bool>visited;
        
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = q.front();
                q.pop();
                if (visited[node]) continue;
                if (node != v && !node->left && !node->right)
                    ans++;
                visited[node] = true;
                for (auto j = 0; j < mp[node].size(); j++)
                    q.push(mp[node][j]);
            }
            dist++;
            if (dist > distance) return;
        }
    }
    
public:
    int countPairs(TreeNode* root, int distance) {
        if (!root) return 0;
        storeLeaves(root);
        for (int i = 0; i < leaves.size(); i++)
            bfs(leaves[i], distance);
        return ans/2;
    }
};
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