Number of Good Leaf Nodes Pairs
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1530.Number%20of%20Good%20Leaf%20Nodes%20Pairs/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1530.Number%20of%20Good%20Leaf%20Nodes%20Pairs/images/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>
Tree; Depth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int countPairs(TreeNode root, int distance) {
        if (root == null) {
            return 0;
        }
        int ans = countPairs(root.left, distance) + countPairs(root.right, distance);
        int[] cnt1 = new int[distance];
        int[] cnt2 = new int[distance];
        dfs(root.left, cnt1, 1);
        dfs(root.right, cnt2, 1);
        for (int i = 0; i < distance; ++i) {
            for (int j = 0; j < distance; ++j) {
                if (i + j <= distance) {
                    ans += cnt1[i] * cnt2[j];
                }
            }
        }
        return ans;
    }

    void dfs(TreeNode root, int[] cnt, int i) {
        if (root == null || i >= cnt.length) {
            return;
        }
        if (root.left == null && root.right == null) {
            ++cnt[i];
            return;
        }
        dfs(root.left, cnt, i + 1);
        dfs(root.right, cnt, i + 1);
    }
}