Maximum Level Sum of a Binary Tree

Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.

Return the smallest level x such that the sum of all the values of nodes at level x is maximal.

Example 1:

Input: root = [1,7,0,7,-8,null,null]
Output: 2
Explanation: 
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.

Example 2:

Input: root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output: 2

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -105 <= Node.val <= 105

思路：就是个level traverse。

/**
 * 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 int maxLevelSum(TreeNode root) {
        if(root == null) {
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        int globalsum = Integer.MIN_VALUE;
        int maxlevel = 0;
        int level = 1;
        while(!queue.isEmpty()) {
            int size = queue.size();
            int cursum = 0;
            for(int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                cursum += node.val;
                if(node.left != null) {
                    queue.offer(node.left);
                }
                if(node.right != null) {
                    queue.offer(node.right);
                }
            }
            if(cursum > globalsum) {
                globalsum = cursum;
                maxlevel = level;
            }
            level++;
        }
        return maxlevel;
    }
}