题解 | #重量级的一层#

题目考察的知识点 考察二叉树层序遍历

问题分析： 只需层序遍历二叉树,计算当前层数重量与目前最大重量比较即可

本题所用的编程语言:Java

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
        if (root == null) return 1;
        Queue<TreeNode> queue = new LinkedList<>();
        TreeNode temp;
        int height = 1;
        int indexH = height;
        int weight = 0;
        int maxWeight = 0;
        queue.add(root);
        while (!queue.isEmpty()) {
            weight = 0;
            for (TreeNode node : queue) {
                weight += node.val;
            }
            if(weight >= maxWeight) {
               height = indexH; 
               maxWeight = weight;
            }
            int size = queue.size();
            while (size-- > 0) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            indexH++;
        }
        return height;
    }

}