<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:

    set<pair<TreeNode*, TreeNode*>>se;
    unordered_map<TreeNode*, TreeNode*>par;
    unordered_set<TreeNode*>vis;

    void dfs(TreeNode* curr, TreeNode* src, int d, int k) {

        if(!curr || d > k || vis.find(curr) != vis.end()) {
            return;
        }

        vis.insert(curr);
        if(d > 0 && d <= k && !curr->left && !curr->right) {
            if(se.find({src, curr}) == se.end() && se.find({curr, src}) == se.end()) {
                se.insert({src, curr});
            }
        }
        dfs(par[curr], src, d+1, k);
        dfs(curr->left, src, d+1, k);
        dfs(curr->right, src, d+1, k);
    }

    int countPairs(TreeNode* root, int distance) {

        se = set<pair<TreeNode*, TreeNode*>>();
        par = unordered_map<TreeNode*, TreeNode*>();

        queue<TreeNode*>q;
        q.push(root);

        queue<TreeNode*>nodes;

        while(!q.empty()) {

            TreeNode* curr = q.front(); q.pop();
            if(!curr->left && !curr->right) {
                nodes.push(curr);
            }

            if(curr->left) {
                par[curr->left] = curr;
                q.push(curr->left);
            }
            if(curr->right) {
                par[curr->right] = curr;
                q.push(curr->right);
            }
        }

        while(!nodes.empty()) {

            vis = unordered_set<TreeNode*>();
            TreeNode* curr = nodes.front(); nodes.pop();
            dfs(curr, curr, 0, distance);
        }
        return se.size();
    }
};
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