Problem
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, 10^4].-10^5 <= Node.val <= 10^5
Solution (Java)
/**
* 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 {
private List<Integer> sums;
public int maxLevelSum(TreeNode root) {
sums = new ArrayList<>();
find(root, 1);
int ans = 1;
int maxv = Integer.MIN_VALUE;
for (int i = 0; i < sums.size(); i++) {
if (sums.get(i) > maxv) {
maxv = sums.get(i);
ans = i + 1;
}
}
return ans;
}
private void find(TreeNode root, int height) {
if (root == null) {
return;
}
if (sums.size() < height) {
sums.add(root.val);
} else {
sums.set(height - 1, sums.get(height - 1) + root.val);
}
find(root.left, height + 1);
find(root.right, height + 1);
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).