1161. Maximum Level Sum of a Binary Tree.java

M
tags: Graph, BFS, DFS
time: O(n) visit all nodes
space: O(n)


#### BFS
- simply calc each level sum with BFS
- top-level is processed first, since we go from top level -> deeper level
    - only update result if sum is truly > global MAX.

```
/*
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: [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.
 

Note:

The number of nodes in the given tree is between 1 and 10^4.
-10^5 <= node.val <= 10^5
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max = Integer.MIN_VALUE;
    
    public int maxLevelSum(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int result = 0, level = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();
            int sum = 0;
            level++;
            while (size-- > 0) {
                TreeNode node = queue.poll();
                sum += node.val;
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            if (sum > max) {
                max = sum;
                result = level;
            }
        }
        
        return result;
    }
}
```