题解 | #重量级的一层#

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    public int maxLevelSum (TreeNode root) {
        // write code here
        List<Long> list = new ArrayList<>();
        levelOrder(root, 0, list);
        int size = list.size(), ans = size;
        long max = list.get(size - 1);
        for (int i = size - 2; i >= 0; i--) {
            long x = list.get(i);
            if (x > max) {
                max = x;
                ans = i + 1;
            }
        }
        return ans;
    }
    private void levelOrder(TreeNode root, int level, List<Long> list) {
        if (root == null) return;
        if (level == list.size()) list.add(0L);
        list.set(level, list.get(level) + root.val);
        levelOrder(root.left, level + 1, list);
        levelOrder(root.right, level + 1, list);
    }
}

• 利用二叉树的层序遍历，计算二叉树每一层节点数值的累加和，将每一层的累加和存储到一个List集合中
• 逆序遍历List集合，确定最大值以及所在层