Number of Good Leaf Nodes Pairs

Tested Python solution for LeetCode 1530 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 1530, Medium. Topics: Tree, Depth-First Search, Binary Tree, DP on Trees. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1530   # by problem number
lcpy gen -s number_of_good_leaf_nodes_pairs   # by problem name

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Problem

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.

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Examples

Input: root = [1,2,3,null,4], distance = 3
Output: 1

Explanation: 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.

Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2

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.

Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1

Explanation: The only good pair is [2,5].

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Constraints

• The number of nodes in the tree is in the range [1, 2^10].
• 1 <= Node.val <= 100
• 1 <= distance <= 10

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from leetcode_py import TreeNode


class 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

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Complexity

Time	Space
O(n * d^2) where d = distance (each leaf-depth list holds at most d entries)	O(n)

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Tags

NeetCode All.