<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 N;
    map<int, int>is_leaf_node;
    vector<vector<int>>adj_list;

    void DFS(TreeNode* root){

        int u = root->val - 1, l, r;
        if(root->left == NULL && root->right == NULL){

            is_leaf_node[u]++;
            return;
        }

        if(root->left){

            l = root->left->val - 1;
            adj_list[u].push_back(l);
            adj_list[l].push_back(u);
            DFS(root->left);
        }

        if(root->right){
            r = root->right->val - 1;
            adj_list[u].push_back(r);
            adj_list[r].push_back(u);
            DFS(root->right);
        }
    }

    void preProcess(TreeNode* root, int i){

        int sz = 1, val = 1;
        TreeNode* u;
        queue<TreeNode*>q;
        q.push(root);

        while(sz){

            for(int i = 0; i < sz; i++){
                u = q.front();
                q.pop();
                u->val = val++;
                if(u->left) q.push(u->left);
                if(u->right) q.push(u->right);
            }
            sz = q.size();
        }
        N = val;
    }

    void BFS(int src, int& ans, int distance){

        queue<int>q;
        q.push(src);
        vector<int>visited(N, 0);

        int level = 1, sz = 1, u;

        while(!q.empty() && level <= distance){
            for(int i = 0; i < sz; i++){

                u = q.front();
                q.pop();
                visited[u] = true;

                for(auto v: adj_list[u]){
                    if(!visited[v]){
                        if(is_leaf_node[v]) ans++;
                        q.push(v);
                    }
                }
            }

            sz = q.size();
            level++;
        }
    }


    int countPairs(TreeNode* root, int distance) {

        preProcess(root, 1);
        adj_list.resize(N);
        DFS(root);

        int ans = 0;
        for(auto u: is_leaf_node){
            if(is_leaf_node[u.first])
                BFS(u.first, ans, distance);
        }

        return ans/2;
    }
};
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