<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>

class Solution {
    /* Step to solve 
    1. in vectore<TreeNode> store all the  root node TC: O(n)  SC: O(n)  max rootnode = 2^10+1/2 = 1024+1/2 = 512
    2) convert it intro set for faster lookup TC: O(n) SC: O(n)
    3) convert it into graph TC: O(n) SC: O(n)
    4) do dfs to find sortest distance bw root node but if length in >  distance break  Tc: 10*rootnode

    overall time complexity: 1 + 2+ 4  = O(n)
    */
public:
    int countPairs(TreeNode* root, int distance) {
        unordered_set<TreeNode*> s;
        int ans = 0;
        unordered_map<TreeNode*, vector<TreeNode*>> graph;
        fill(root, s, graph);
        for(auto start : s){
            unordered_set<TreeNode*> visi;
            dfs(start,s, graph, distance, ans,visi);    
        }
        
        return ans/2;

    }
    // TC:O(n)
    void fill(TreeNode* root, unordered_set<TreeNode*>& s , unordered_map<TreeNode*, vector<TreeNode*>>& graph){
        if(root == nullptr){
            return;
        }
        if(root->left == nullptr && root->right == nullptr){
            s.insert(root);
        }

        if(root->left != nullptr){
            graph[root].push_back(root->left);
            graph[root->left].push_back(root);
            fill(root->left, s, graph);
        }
        if(root->right != nullptr){
            graph[root].push_back(root->right);
            graph[root->right].push_back(root);
            fill(root->right, s, graph);
        }


    }

    void dfs(TreeNode* curr,unordered_set<TreeNode*>& s, unordered_map<TreeNode*, vector<TreeNode*>>& graph, int distance, int& ans,unordered_set<TreeNode*>& visited){
        if(distance  <1){
            return;
        }
        // visited set
        // cout<<"current distance is = "<<distance<<" curr node is "<<curr->val<<" ";
        visited.insert(curr);
        for(auto next : graph[curr]){
            if(visited.find(next) == visited.end()){
                // cout<<"nbr is "<<next->val<<" ";
                if(s.find(next) != s.end()){
                    // cout<<"!!!!find answer at distance :"<<distance<<" node is "<<next->val;
                    ans++;
                }
                dfs(next, s, graph, distance-1, ans, visited);
            }
        }
        // cout<<endl;

    }

private:

};
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