<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 goodNodes(TreeNode* root,unordered_map<TreeNode*,vector<TreeNode*>>&adj,int dis)
{
    unordered_map<TreeNode*,int>vis;
    queue<TreeNode*>q;
    q.push(root);
    int ans=0;
    while(q.size()>0 && dis>0)
    {
        int sz=q.size();
        for(int i=0;i<sz;i++)
        {
            TreeNode* head=q.front();
            vis[head]=1;
            q.pop();
            for(auto it:adj[head])
            {
                if(!vis[it])
                {
                    if(!it->left && !it->right) ans++;
                    q.push(it);
                }
            }
        }
        dis--;
    }
    // while(q.size()>0)
    // {
    //     TreeNode* head=q.front();
    //     q.pop();
    //     if(!head->left && !head->right) ans++;
    // }
    return ans;
}
void dfs(TreeNode* root,unordered_map<TreeNode*,vector<TreeNode*>>&adj)
{
    if(!root)return;
    if(root->left)
    {
        adj[root->left].push_back(root);
        adj[root].push_back(root->left);
    }
    if(root->right)
    {
        adj[root->right].push_back(root);
        adj[root].push_back(root->right);
    }
    dfs(root->left,adj);
    dfs(root->right,adj);
}
int dfs2(TreeNode* root,unordered_map<TreeNode*,vector<TreeNode*>>&adj,int dis)
{
    if(!root)return 0;
    if(!root->left && !root->right)
    {
        return goodNodes(root,adj,dis);
    }
    return dfs2(root->left,adj,dis)+dfs2(root->right,adj,dis);
}
    int countPairs(TreeNode* root, int distance) {
     if(distance==1)return 0;
     unordered_map<TreeNode*,vector<TreeNode*>>adj;
     dfs(root,adj);
     return dfs2(root,adj,distance)/2;
    }
};
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