<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) {}
 * };
 */


/**
 * 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:
// void solve(TreeNode* root, vector<TreeNode*>&leaf){
//     if(root != NULL && root->left == NULL && root->right == NULL){
//         leaf.push_back(root);
//         return ;
//     }
//     if(root == NULL){
//         return ;
//     }
//     solve(root->left,leaf);
//     solve(root->right,leaf);
// }
// TreeNode* comman(TreeNode* root, TreeNode* one,TreeNode* two){
//     if(root == NULL){
//         return NULL;
//     }
//     if(root == one){
//         return one;
//     }
//     if(root == two){
//         return two;
//     }
//     TreeNode* lef = comman(root->left,one,two);
//     TreeNode* rig = comman(root->right,one,two);
//     if(lef != NULL && rig == NULL){
//         return lef;
//     }
//     if(lef == NULL && rig != NULL){
//         return rig;
//     }
//     if(lef != NULL && rig != NULL){
//         return root;
//     }
//     else{
//         return NULL;
//     }
// }

//  void dis(TreeNode* prnt,TreeNode* one,TreeNode* two,int right,int left,int &l,int &r){
//     if(prnt == NULL){
//         return ;
//     }
//     if(prnt == one){
//         l = left;
//     }
//      if(prnt == two){
//         r = left;
//     }
//     dis(prnt->left,one,two,right,left+1,l,r);
//     dis(prnt->right,one,two,right,left+1,l,r);

// }

//     int countPairs(TreeNode* root, int distance) {
//         vector<TreeNode*>leaf;
//         solve(root,leaf);
//         int ans =0;
//        for(int i =0;i<leaf.size();i++){
//           for(int j = i+1;j<leaf.size();j++){
//             TreeNode* one = leaf[i];
//             TreeNode* two = leaf[j];
//            TreeNode * prnt = comman(root,one,two);
//            int l =0;
//            int r = 0;
//            dis(prnt,one,two,0,0,l,r);
//            int d = l+r;

//            if(d<=distance){
//            ans++;
//            }
//           }
//        }
//        return ans;
      
//     }
// };
//ONE WITH FINDING THE LCA and then 

//now just store the string path to leaf
class Solution {
public:
void solve(TreeNode* root,string &str,vector<string>&leaf){
    if(root == NULL){
        return ;
    }
    if(root->left == NULL && root->right == NULL){
        leaf.push_back(str);
        return ;
    }

    if(root->left){
        string prev = str;
        str = str + "l";
        

        solve(root->left,str,leaf);
        str = prev;
    }
    if(root->right){
          string prev = str;
        str = str + "r";
        

        solve(root->right,str,leaf);
        str = prev;
    }

}
    int countPairs(TreeNode* root, int distance) {
        string str = "";
        vector<string>leaf;
        solve(root,str,leaf);
        int ans =0 ;
        for(int i =0;i<leaf.size();i++){
            for(int j = i+1;j<leaf.size();j++){
                string one = leaf[i];
                 string two = leaf[j];

                int cmn = 0;
                while(cmn<one.length() && cmn<two.length()){
                    if(one[cmn]!= two[cmn]){
                        break;
                    }
                    else{
                        cmn++;
                    }
                }


                int diff = one.length() - cmn + two.length() - cmn;
                if(diff<=distance){
                    ans++;
                }
            }
        }
        return ans;
    }
};
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