<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 ans = 0;
    int countPairs(TreeNode* root, int distance) {
        // Build A Graph O(N)
        // and get a list of all leaf nodes
        unordered_map<TreeNode*, vector<TreeNode*>> G;
        unordered_set<TreeNode*> leaf;
        deque<TreeNode*> q;
        q.push_back(root);
        while(!q.empty()){
            TreeNode* cur = q.front();
            q.pop_front();
            if(!cur->left && !cur->right){
                leaf.insert(cur);
            }
            if(cur->left){
                q.push_back(cur->left);
                G[cur].push_back(cur->left);
                G[cur->left].push_back(cur);
            }
            if(cur->right){
                q.push_back(cur->right);
                G[cur].push_back(cur->right);
                G[cur->right].push_back(cur);
            }
        }
        // iterate over all the good node pairs
        for(TreeNode* s: leaf){
            int curDistance = 0;
            deque<TreeNode*> q2;
            unordered_set<TreeNode*> visited;
            q2.push_back(s);
            visited.insert(s);
            while(!q2.empty()){
                int q2Size = q2.size();
                for(int cnt = 0; cnt < q2Size; ++cnt){
                    TreeNode* u = q2.front();
                    q2.pop_front();
                    for(TreeNode* v: G[u]){
                        if(visited.find(v) == visited.end()){
                            visited.insert(v);
                            q2.push_back(v);
                            if(curDistance+1 <= distance && leaf.find(v)!=leaf.end()){
                                ans ++;
                            }
                        }
                    }
                }
                curDistance ++;
                if(curDistance >= distance){
                    break;
                }
            }
        }
        return ans/2;
    }
};
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