<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 func(TreeNode*root,int dis,unordered_map<TreeNode*,TreeNode*>& parent){
        queue<TreeNode*> q;
        unordered_map<TreeNode*,bool> vis;
        q.push(root);
        int k = 0;
        int cnt = 0;
        while(k<=dis){
            int siz = q.size();
            while(siz > 0){
                TreeNode* temp = q.front();
                if(temp->left == NULL && temp->right == NULL && temp != root){
                    cnt++;
                }
                q.pop();
                vis[temp] = true;
                //push its parents and childs if not visited
                if(temp->left && vis[temp->left]==false)
                q.push(temp->left);
                if(temp->right && vis[temp->right] == false)
                q.push(temp->right);
                if(parent[temp] && vis[parent[temp]]==false)
                q.push(parent[temp]);
                siz--;
            }
            k++;
        }
        return cnt;
    }
    void rec(TreeNode* curr,TreeNode*prev,unordered_map<TreeNode*,TreeNode*>& par){
        if(curr == NULL) return;
        par[curr] = prev;
        rec(curr->left,curr,par);
        rec(curr->right,curr,par);
        return;
    }
    int rec2(TreeNode* root,int dis,unordered_map<TreeNode*,TreeNode*>& par){
        if(root == NULL) return 0;
        int count = 0;
        if(root->left == NULL && root->right == NULL){
            //its a leaf node 
            count = func(root,dis,par);
        }
        int lef = rec2(root->left,dis,par);
        int ri = rec2(root->right,dis,par);
        return count + lef + ri;
    }
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*,TreeNode*> par;
        rec(root,NULL,par);
        //get all the leaf nodes and call for the fucntion
        int ans = rec2(root,distance,par);
        return ans/2;
    }
};
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