<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>

class Solution {
    vector<TreeNode*> leaves;
    unordered_map<TreeNode* , TreeNode*> pof;
    void makeGraph(unordered_map<TreeNode* , TreeNode*> &pof, TreeNode* root){
        if(!root) return;
        if(!root->left && !root->right){
            leaves.push_back(root);
            return;
        }
        if(root->left) pof[root->left]=root;
        if(root->right) pof[root->right]=root;
        makeGraph(pof, root->left);
        makeGraph(pof, root->right);
    }
public:
    int countPairs(TreeNode* root, int distance) {
        leaves={};
        makeGraph(pof, root);
        int ans=0;
        queue<pair<TreeNode*, int>> q;
        for(auto &l: leaves){
            q.push({l, 0}); 
            unordered_set<TreeNode*> isVis;
            while(q.size()){
                auto [cur, dist]=q.front();
                q.pop();
                if(isVis.count(cur) || dist>distance) continue;
                isVis.insert(cur);
                if(!cur->left && !cur->right && cur!=l && dist<=distance){
                    ans++;
                    continue;
                }
                if(cur->left && !isVis.count(cur->left)) q.push({cur->left, dist+1});
                if(cur->right && !isVis.count(cur->right)) q.push({cur->right, dist+1});
                if(pof.count(cur) && !isVis.count(pof[cur])) q.push({pof[cur], dist+1});
            }
        }
        return ans>>1;
    }
};
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