<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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def countPairs(self, root: TreeNode, distance: int) -> int:
        graph = {}
        leafNodes = set()
        self.convertTree(root, graph, leafNodes)
        self.res = 0
        queue = []
        for node in leafNodes:
            queue.append([node, 0])
            self.traverseGraph(queue, node, graph, distance, leafNodes)
        return self.res // 2

    def convertTree(self, node, graph, leafNodes):
        if node is None:
            return
        if node not in graph:
            graph[node] = set()
        if node.left:
            graph[node].add(node.left)
            if node.left not in graph:
                graph[node.left] = set()
            graph[node.left].add(node)
        if node.right:
            graph[node].add(node.right)
            if node.right not in graph:
                graph[node.right] = set()
            graph[node.right].add(node)
        if node.right is None and node.left is None:
            leafNodes.add(node)
        self.convertTree(node.left, graph, leafNodes)
        self.convertTree(node.right, graph, leafNodes)

    def traverseGraph(self, queue, start, graph, maxDist, leafNodes):
        seen = set()
        while (len(queue) > 0):
            [node, curDist] = queue.pop(0)
            if start != node and node not in seen and node in leafNodes and curDist <= maxDist:
                seen.add(node)
                self.res += 1
            if curDist + 1 > maxDist:
                continue
            for directedNode in graph[node]:
                queue.append([directedNode, curDist + 1])

18000