<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 {
public:
    int countPairs(TreeNode* root, int distance) {
        total = 0;
        dfs(root, distance);
        return total;
    }

    int total;

    vector<int> dfs(TreeNode* root, int dist) {
        if (root == nullptr) {
            return {};
        }

        auto l = dfs(root->left, dist);
        auto r = dfs(root->right, dist);
        vector<int> res;

        if (l.empty() && r.empty()) {
            return { 0 };
        } else {
            int sum = 0;
            int i = 0, j = r.size() - 1;
            while (i < l.size() && j >= 0) {
                if (l[i] + r[j] + 2 <= dist) {
                    sum += (j + 1);
                    ++i;
                } else {
                    --j;
                }
            }
            total += sum;
        }

        int i = 0, j = 0;
        while (i < l.size() && j < r.size()) {
            if (l[i] <= r[j]) {
                res.push_back(l[i] + 1);
                ++i;
            } else {
                res.push_back(r[j] + 1);
                ++j;
            }
        }
        while (i < l.size() ) {
            res.push_back(l[i] + 1);
            ++i;
        }
        while (j < r.size() ) {
            res.push_back(r[j] + 1);
            ++j;
        }

        return res;
    }
};
37006