Kth Largest Sum in a Binary Tree

Tested Python solution for LeetCode 2583 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 2583, Medium. Topics: Tree, Breadth-First Search, Sorting, Binary Tree. View on LeetCode.

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

lcpy gen -n 2583   # by problem number
lcpy gen -s kth_largest_sum_in_a_binary_tree   # by problem name

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Problem

You are given the root of a binary tree and a positive integer k.

The level sum in the tree is the sum of the values of the nodes that are on the same level.

Return the kth largest level sum in the tree (not necessarily distinct). If there are fewer than k levels in the tree, return -1.

Note that two nodes are on the same level if they have the same distance from the root.

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Examples

Input: root = [5,8,9,2,1,3,7,4,6], k = 2
Output: 13
Explanation: The level sums are the following:
- Level 1: 5.
- Level 2: 8 + 9 = 17.
- Level 3: 2 + 1 + 3 + 7 = 13.
- Level 4: 4 + 6 = 10.
The 2nd largest level sum is 13.

Input: root = [1,2,null,3], k = 1
Output: 3
Explanation: The largest level sum is 3.

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Constraints

• The number of nodes in the tree is n.
• 2 <= n <= 10^5
• 1 <= Node.val <= 10^6
• 1 <= k <= n

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Solution

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

from collections import deque

from leetcode_py import TreeNode


class Solution:
    # Time: O(n)
    # Space: O(w)
    def kth_largest_level_sum(self, root: TreeNode[int] | None, k: int) -> int:
        sums: list[int] = []
        queue = deque([root] if root is not None else [])
        while queue:
            level_sum = 0
            for _ in range(len(queue)):
                node = queue.popleft()
                level_sum += node.val
                if node.left is not None:
                    queue.append(node.left)
                if node.right is not None:
                    queue.append(node.right)
            sums.append(level_sum)
        if k > len(sums):
            return -1
        sums.sort(reverse=True)
        return sums[k - 1]

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Complexity

Time	Space
O(n)	O(w)

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Tags

NeetCode All.