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:**
1. The number of nodes in the given tree is between `1` and `10^4`.
2. `-10^5 <= node.val <= 10^5`
暴力解法：
public int maxLevelSum(TreeNode root) {
int[] sum = new int[40];
int level = 0;
check(root,sum,level);
int retsum = 0;
int retlevel = 0;
for(int i=0;i<sum.length;i++){
if(retsum<sum[i]){
retsum = sum[i];
retlevel = i;
}
}
return retlevel+1;
}
public void check(TreeNode root,int[] sum,int level){
sum[level] = sum[level] +root.val;
if(root.left!=null){
check(root.left,sum,level+1);
}
if(root.right!=null){
check(root.right,sum,level+1);
}
}
目前题目提交的人还是太少了，只有6千多通过，暴力都通过。
Runtime: 3 ms, faster than 100.00% of Java online submissions for Maximum Level Sum of a Binary Tree.
Memory Usage: 41.3 MB, less than 100.00% of Java online submissions for Maximum Level Sum of a Binary Tree.
