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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func countPairs(root *TreeNode, distance int) int {
	if root == nil {
		return 0
	}
	ans := countPairs(root.Left, distance) + countPairs(root.Right, distance)
	cnt1 := make([]int, distance)
	cnt2 := make([]int, distance)
	dfs(root.Left, cnt1, 1)
	dfs(root.Right, cnt2, 1)
	for i, v1 := range cnt1 {
		for j, v2 := range cnt2 {
			if i+j <= distance {
				ans += v1 * v2
			}
		}
	}
	return ans
}

func dfs(root *TreeNode, cnt []int, i int) {
	if root == nil || i >= len(cnt) {
		return
	}
	if root.Left == nil && root.Right == nil {
		cnt[i]++
		return
	}
	dfs(root.Left, cnt, i+1)
	dfs(root.Right, cnt, i+1)
}