<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>

// 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 traverse(TreeNode* root, map<TreeNode*, vector<TreeNode*> >& g, set<TreeNode*>& leaves){
        if(!root)
            return;
        if(root->right){
            g[root].push_back(root->right);
            g[root->right].push_back(root);
        }
        if(root->left){
            g[root].push_back(root->left);
            g[root->left].push_back(root);
        }
        if(!root->left && !root->right){
            leaves.insert(root);
        }
        traverse(root->left, g, leaves);
        traverse(root->right, g, leaves);
    }

    void dfs(TreeNode* root, TreeNode* parent, map<TreeNode*, vector<TreeNode*> >& g, int& ans, int level){
        if(!root || level < 0){
            return ;
        }

        if(!root->left && !root->right && parent != nullptr){
            // cerr << "    Good: " << root->val << ", level: " << level << endl;
            ans++;
        }

        // cerr << "  IN: " << root->val << ", level: " << level << endl;
        for(TreeNode* neigh : g[root]){
            if(neigh == parent)
                continue;
            // cerr << "  DFS: " << neigh->val << endl;
            dfs(neigh, root, g, ans, level - 1);
        }
    }
public:
    int countPairs(TreeNode* root, int distance) {
        map<TreeNode*, vector<TreeNode*> > g;
        set<TreeNode*> leaves;
        traverse(root, g, leaves);
        int ans = 0;
        for(TreeNode* leaf : leaves){
            // DFS Distance
            // cerr << "DFS: " << leaf->val << endl;
            dfs(leaf, nullptr, g, ans, distance);
        }

        assert(ans % 2 == 0);
        // cerr << "Ans: " << ans << endl;
        return ans / 2;
    }
};
44318