You are given the root of a binary tree and an integer distance.A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance.Return the number of good leaf node pairs in the tree.
Explanation: The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of the shortest path between them is 4.
from leetcode_py import TreeNodeclass Solution: # Time: O(n * d^2) where d = distance (each leaf-depth list holds at most d entries) # Space: O(n) def count_pairs(self, root: TreeNode[int] | None, distance: int) -> int: total = 0 def dfs(node: TreeNode[int] | None) -> list[int]: nonlocal total if node is None: return [] if node.left is None and node.right is None: return [1] left = dfs(node.left) right = dfs(node.right) for left_depth in left: for right_depth in right: if left_depth + right_depth <= distance: total += 1 return [depth + 1 for depth in left + right if depth + 1 < distance] dfs(root) return total
