<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:
    vector<TreeNode*>leaf;
    void getLeaf(TreeNode *root, unordered_map<TreeNode*, TreeNode*>&parent, TreeNode *parent1) {
        if (root == NULL) return;
        parent[root] = parent1;
        if ((root->left == NULL) && (root->right == NULL)) {
            leaf.push_back(root);
        }
        getLeaf(root->left, parent, root);
        getLeaf(root->right, parent, root);
    }
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*, TreeNode*>mp;
        getLeaf(root, mp, NULL);
        int count = 0;
        for(int i = 0; i < leaf.size(); ++i) {
            queue<pair<TreeNode*, int>>q;
            q.push({leaf[i], 0});
            unordered_set<TreeNode*>s;
            s.insert(leaf[i]);
            while(!q.empty()) {
                pair<TreeNode*, int>top = q.front();
                q.pop();
                if ((top.first->left == NULL) && (top.first->right == NULL) && (top.first != leaf[i])) ++count;
                // cout<<top.first->val<<" "<<top.second<<endl;
                if (top.second + 1 > distance) continue;
                if (mp[top.first] != NULL && !s.count(mp[top.first])) {
                    s.insert(mp[top.first]);
                    q.push({mp[top.first], top.second + 1});
                }
                if ((top.first->left != NULL) && (!s.count(top.first->left))) {
                    s.insert(top.first->left);
                    q.push({top.first->left, top.second + 1});
                }
                if ((top.first->right != NULL) && (!s.count(top.first->right))) {
                    s.insert(top.first->right);
                    q.push({top.first->right, top.second + 1});
                }
            }
        }
        return count / 2;
    }
};
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