Breadth-First Search
Level Order Sum
easy
Count: 10
DESCRIPTION
Given the root of a binary tree, return the sum of the nodes at each level. The output should be a list containing the sum of the nodes at each level.
Example 1:
Input:
[1, 3, 4, null, 2, 7, null, 8]
Output: [1, 7, 9, 8]
Example 2:
Input:
[1, 2, 5, 3, null, null, null, null, 4]
Output: [1, 7, 3, 4]
Explanation
We should recognize that a level-order breadth-first traversal of the binary tree is the most straightforward way to solve this problem.
At each level, we can keep a running sum of the node's values at that level. Then, whenever we finish processing a level (the for-loop for that level finishes), then we can add the sum of the nodes to the output list.
Solution
class Solution:def levelSum(self, root: TreeNode) -> List[int]:if not root:return []nodes = []queue = deque([root])while queue:# start of a new level herelevel_size = len(queue)sum_ = 0# process all nodes in the current levelfor i in range(level_size):node = queue.popleft()sum_ += node.valif node.left:queue.append(node.left)if node.right:queue.append(node.right)# we are at the end of the level,# add the sum of the nodes to the output listnodes.append(sum_)return nodes
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Complexity Analysis
Time Complexity: O(N) where N is the number of nodes in the tree. We visit each node exactly once, and each node, we perform a constant amount of work.
Space Complexity: O(N) In the worst case, each node is on its own level, so the output list will contain N elements.
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