<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 {
public:
    int ans = 0;
    unordered_map<TreeNode*, TreeNode*> mp;
    void dfs1(TreeNode *root, TreeNode *parent, vector<TreeNode*> &leaf_nodes) {
        if (root == NULL) {
            return;
        }

        if (parent != NULL) {
            mp[root] = parent;
        }

        if (root -> left == NULL && root -> right == NULL) {
            leaf_nodes.push_back(root);
            return;
        }

        dfs1(root->left, root, leaf_nodes);
        dfs1(root->right, root, leaf_nodes);
    }

    void dfs2(TreeNode *root,int depth,int distance, unordered_map<TreeNode*, bool> &vis) {
        if (root == NULL || vis[root] || depth > distance) 
            return;
        
        vis[root] = true;

        if (root -> left == NULL && root -> right == NULL && depth > 0 && depth <= distance) ans++;

        if (mp.find(root) != mp.end() && vis.find(mp[root]) == vis.end())
            dfs2(mp[root], depth + 1, distance, vis);

        if (vis.find(root -> left) == vis.end())
            dfs2(root -> left, depth + 1, distance, vis);

        if (vis.find(root -> right) == vis.end())
            dfs2(root -> right, depth + 1, distance, vis);
    }
    
    int countPairs(TreeNode* root, int distance) {
        vector<TreeNode*> leaf_nodes;
        dfs1(root, NULL, leaf_nodes);

        for (auto leaf_node : leaf_nodes) {
            unordered_map<TreeNode*, bool> vis;
            dfs2(leaf_node, 0, distance, vis);
        }

        return ans / 2;
    }
};
101356