<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:
    vector<TreeNode*> leaves;
    unordered_map<TreeNode*, TreeNode*> m;
    void getLeafNodes(TreeNode* root) {
        if (root == NULL) {
            return;
        }
        if (root->left == NULL && root->right == NULL) {
            leaves.push_back(root);
        }
        getLeafNodes(root->left);
        getLeafNodes(root->right);
    }
    void getParentNode(TreeNode* root) {
        if (root == NULL) {
            return;
        }
        if (root->left != NULL) {
            m[root->left] = root;
        }
        if (root->right != NULL) {
            m[root->right] = root;
        }
        getParentNode(root->left);
        getParentNode(root->right);
    }
    void func(TreeNode* leafNode, int& count, int distance) {
        queue<pair<TreeNode*, int>> q;
        unordered_map<TreeNode*, bool> visited;

        q.push({leafNode, 0});
        visited[leafNode] = true;
        while (!q.empty()) {
            int size = q.size();
            int flag = 0;
            while (size--) {
                pair<TreeNode*, int> p = q.front();
                q.pop();

                TreeNode* currNode = p.first;
                int d = p.second;

                if (d > distance) {
                    return;
                }

                if (currNode->left == NULL && currNode->right == NULL &&
                    d <= distance && currNode != leafNode) {
                    count++;
                }

                if (currNode->left != NULL && !visited[currNode->left]) {
                    q.push({currNode->left, d + 1});
                    visited[currNode->left] = true;
                }
                if (currNode->right != NULL && !visited[currNode->right]) {
                    q.push({currNode->right, d + 1});
                    visited[currNode->right] = true;
                }
                if (m.find(currNode) != m.end() && !visited[m[currNode]]) {
                    q.push({m[currNode], d + 1});
                    visited[m[currNode]] = true;
                }
            }
        }
    }
public:
    int countPairs(TreeNode* root, int distance) {
        getLeafNodes(root);
        getParentNode(root);
        int count = 0;

        for (int i = 0; i < leaves.size(); i++) {
            func(leaves[i], count, distance);
        }
        return count / 2;
    }
};
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