[LeetCode] 2583. Kth Largest Sum in a Binary Tree

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.

Example 1:

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.

Example 2:

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

Constraints:
The number of nodes in the tree is n.
2 <= n <= 105
1 <= Node.val <= 106
1 <= k <= n

二叉树中的第 K 大层和。

给你一棵二叉树的根节点 root 和一个正整数 k 。
树中的 层和 是指 同一层 上节点值的总和。
返回树中第 k 大的层和（不一定不同）。如果树少于 k 层，则返回 -1 。
注意，如果两个节点与根节点的距离相同，则认为它们在同一层。

思路

简单 BFS 题，用一个 list 存每一层的节点值的和，然后返回第 k 大的，如果树少于 k 层，则返回 -1 。

复杂度

时间O(n)
空间O(n)

代码

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public long kthLargestLevelSum(TreeNode root, int k) {
        List<Long> list = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            long sum = 0L;
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                sum += cur.val;
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            list.add(sum);
        }
        if (list.size() < k) {
            return -1;
        }
        int n = list.size();
        Collections.sort(list);
        return list.get(n - k);
    }
}