<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>

class Solution:
    def _post_order(self, current_node, distance):
        if current_node is None:
            return [0] * 12
        elif current_node.left is None and current_node.right is None:
            current = [0] * 12
            # Leaf node's distance from itself is 0
            current[0] = 1
            return current

        # Leaf node count for a given distance i
        left = self._post_order(current_node.left, distance)
        right = self._post_order(current_node.right, distance)

        current = [0] * 12

        # Combine the counts from the left and right subtree and shift by
        # +1 distance
        for i in range(10):
            current[i + 1] += left[i] + right[i]

        # Initialize to total number of good leaf nodes pairs from left and right subtrees.
        current[11] = left[11] + right[11]

        # Iterate through possible leaf node distance pairs
        for d1 in range(distance + 1):
            for d2 in range(distance + 1):
                if 2 + d1 + d2 <= distance:
                    # If the total path distance is less than the given distance limit,
                    # then add to he total number of good pairs
                    current[11] += left[d1] * right[d2]

        return current

    def countPairs(self, root: TreeNode, distance: int) -> int:
        return self._post_order(root, distance)[11]
17000