Maximum Level Sum of a Binary Tree
<p>Given the <code>root</code> of a binary tree, the level of its root is <code>1</code>, the level of its children is <code>2</code>, and so on.</p>

<p>Return the <strong>smallest</strong> level <code>x</code> such that the sum of all the values of nodes at level <code>x</code> is <strong>maximal</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1161.Maximum%20Level%20Sum%20of%20a%20Binary%20Tree/images/capture.jpg" style="width: 200px; height: 175px;" />
<pre>
<strong>Input:</strong> root = [1,7,0,7,-8,null,null]
<strong>Output:</strong> 2
<strong>Explanation: </strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [989,null,10250,98693,-89388,null,null,null,-32127]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
/**
 * 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) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        int mx = Integer.MIN_VALUE;
        int i = 0;
        int ans = 0;
        while (!q.isEmpty()) {
            ++i;
            int s = 0;
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.pollFirst();
                s += node.val;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            if (mx < s) {
                mx = s;
                ans = i;
            }
        }
        return ans;
    }
}