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Formatted question description: https://leetcode.ca/all/1161.html
1161. Maximum Level Sum of a Binary Tree (Medium)
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
and10^4
. -10^5 <= node.val <= 10^5
Companies:
Google
Related Topics:
Graph
Solution 1. Level-order Traversal
// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxLevelSum(TreeNode* root) {
if (!root) return 0;
int maxSum = 0, ans = 0, lv = 1;
queue<TreeNode*> q;
q.push(root);
while (q.size()) {
int cnt = q.size(), sum = 0;
while (cnt--) {
auto node = q.front();
q.pop();
sum += node->val;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
if (sum > maxSum) {
maxSum = sum;
ans = lv;
}
++lv;
}
return ans;
}
};
Java
-
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int maxLevelSum(TreeNode root) { if (root == null) return 0; int maxSum = 0; int maxSumLevel = 0; int level = 0; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { level++; int curSum = 0; int size = queue.size(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); curSum += node.val; TreeNode left = node.left, right = node.right; if (left != null) queue.offer(left); if (right != null) queue.offer(right); } if (curSum > maxSum) { maxSum = curSum; maxSumLevel = level; } } return maxSumLevel; } }
-
// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/ // Time: O(N) // Space: O(N) class Solution { public: int maxLevelSum(TreeNode* root) { if (!root) return 0; int maxSum = 0, ans = 0, lv = 1; queue<TreeNode*> q; q.push(root); while (q.size()) { int cnt = q.size(), sum = 0; while (cnt--) { auto node = q.front(); q.pop(); sum += node->val; if (node->left) q.push(node->left); if (node->right) q.push(node->right); } if (sum > maxSum) { maxSum = sum; ans = lv; } ++lv; } return ans; } };
-
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def maxLevelSum(self, root: Optional[TreeNode]) -> int: q = deque([root]) mx = -inf i = 0 while q: i += 1 s = 0 for _ in range(len(q)): node = q.popleft() s += node.val if node.left: q.append(node.left) if node.right: q.append(node.right) if mx < s: mx = s ans = i return ans ############ # 1161. Maximum Level Sum of a Binary Tree # https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/ # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def maxLevelSum(self, root: TreeNode) -> int: level = res = 1 mmax = float('-inf') q = collections.deque([root]) while q: val = 0 n = len(q) for _ in range(n): node = q.popleft() val += node.val for leaf in (node.left, node.right): if leaf: q.append(leaf) if val > mmax: res = level mmax = val level += 1 return res