<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 {
private:
    void findParent(TreeNode* root, unordered_map<TreeNode*, TreeNode*> &parentMap, unordered_set<TreeNode*> &leafNodes)
    {
        if(!root)
            return;
        
        queue<TreeNode*> q;
        
        q.push(root);
        
        while(!q.empty())
        {
            TreeNode* currNode = q.front();
            q.pop();
            
            if(!currNode->left && !currNode->right)
            {   
                leafNodes.insert(currNode);
            }
            
            if(currNode->left)
            {
                q.push(currNode->left);
                parentMap[currNode->left] = currNode;
            }
            
            if(currNode->right)
            {
                q.push(currNode->right);
                parentMap[currNode->right] = currNode;
            }
        }
    }
public:
    int countPairs(TreeNode* root, int distance) {
        if(!root)
            return 0;
        unordered_map<TreeNode*, TreeNode*> parentMap;
        unordered_set<TreeNode*> leafNodes;
        findParent(root, parentMap, leafNodes);
        
        int ans = 0;
        for(auto leaf: leafNodes)
        {
            queue<pair<TreeNode*, int>> q;
            q.push({leaf, 0});             //{node, dist}

            unordered_set<TreeNode*> vis;
            vis.insert(leaf);
            
            while(!q.empty())
            {
                auto curr = q.front();
                q.pop();
                
                TreeNode* node = curr.first;
                int dist = curr.second;
                
                if(dist > distance)
                    continue;
                
                if(!node->left && !node->right) // or if(leafNodes.find(node)!= leafNode.end())
                {
                    if(dist <= distance && dist!=0) //or node != leaf
                        ans++;
                }

                if(parentMap.find(node) != parentMap.end() && vis.find(parentMap[node]) == vis.end())
                {
                    TreeNode* parent = parentMap[node];
                    vis.insert(parent);
                    q.push({parent, dist + 1});
                }
                
                if(node->left && vis.find(node->left) == vis.end())
                {
                    vis.insert(node->left);
                    q.push({node->left, dist+1});
                }
                
                if(node->right && vis.find(node->right) == vis.end())
                {
                    vis.insert(node->right);
                    q.push({node->right, dist+1});
                }
            }
        }
        return ans/2;   //Each pair is counted twice
    }
};

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