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
         Deque<TreeNode> outer = new LinkedList<>();
        Deque<TreeNode> inner = new LinkedList<>();
        if(root != null) outer.offerFirst(root);
        int ans = 0;
        int i = 0;
        int max = 0;
        while (!outer.isEmpty()){
            i++;
            while(!outer.isEmpty()){
                inner.offerFirst(outer.pollLast());
            }
            int num = 0;
            while(!inner.isEmpty()){
                TreeNode treeNode = inner.pollLast();
                num += treeNode.val;
                if(treeNode.left != null) outer.offerFirst(treeNode.left);
                if(treeNode.right != null) outer.offerFirst(treeNode.right);
            }
            if(num >= max) {
                ans = i;
                max = num;
            }
        }
        return ans;
    }
}

知识点:

树，层序遍历

解题思路:

用两个队列层序遍历每一层，用max记录最大值，遍历每一层时用num来累加当前层值，i来记录当前层数。当num大于等于max时，用ans记录当前层数，更新max的值。最总结果就是ans记录的层数。