<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 Bfs(TreeNode* root, int d, unordered_map<TreeNode*, TreeNode*>& mpp) {
        queue<TreeNode*> q;
        unordered_set<TreeNode*> vis;
        q.push(root);
        vis.insert(root);
        int level = 0;
        TreeNode* par = root;
        int cnt = 0;

        while (!q.empty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                TreeNode* node = q.front();
                q.pop();

                if (!node->left && !node->right && node != par) cnt++;

                if (node->left && vis.find(node->left) == vis.end()) {
                    vis.insert(node->left);
                    q.push(node->left);
                }
                if (node->right && vis.find(node->right) == vis.end()) {
                    vis.insert(node->right);
                    q.push(node->right);
                }
                if (mpp.find(node) != mpp.end() && vis.find(mpp[node]) == vis.end()) {
                    vis.insert(mpp[node]);
                    q.push(mpp[node]);
                }
            }
            level++;
            if (level > d) break;
        }

        return cnt;
    }

    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*, TreeNode*> mpp;
        queue<TreeNode*> q;
        vector<TreeNode*> leaf;

        q.push(root);
        while (!q.empty()) {
            TreeNode* node = q.front();
            q.pop();

            if (node->left) {
                mpp[node->left] = node;
                q.push(node->left);
            }
            if (node->right) {
                mpp[node->right] = node;
                q.push(node->right);
            }
            if (!node->left && !node->right) {
                leaf.push_back(node);
            }
        }

        int ans = 0;
        for(auto it:leaf){
            ans+=Bfs(it,distance,mpp);
        }

        return ans/2;
    }
};

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