Is Zigzag Level Sum of Binary Tree easy or hard?

This problem is typically classified as medium difficulty. It requires understanding binary tree traversal, especially BFS level order traversal, and managing level boundaries while maintaining a zigzag traversal pattern.

Zigzag Level Sum of Binary Tree Python/Java solution

Most Python and Java implementations use a queue for BFS. Initialize a queue with the root node, iterate level by level, compute the sum for each level, and toggle a direction flag after each iteration. This produces an O(n) time solution with straightforward code in both languages.

How to solve Zigzag Level Sum of Binary Tree in O(n)?

Perform a BFS traversal using a queue. For each level, determine the number of nodes currently in the queue, process exactly that many nodes, and accumulate their values into a level sum. Push children into the queue for the next level and toggle a direction flag to maintain zigzag order. Since each node is visited once, the algorithm runs in O(n).

What is the best approach for Zigzag Level Sum of Binary Tree?

Breadth-first search (BFS) level order traversal with a queue is the most practical approach. Process the tree one level at a time, sum node values in that level, and alternate traversal direction logically to match the zigzag pattern. This approach visits each node exactly once and runs in O(n) time with O(w) space where w is the maximum width of the tree.

Is Zigzag Level Sum of Binary Tree asked at Google/Amazon/Meta?

Binary tree level-order and zigzag traversal variants frequently appear in interviews at companies like Amazon, Google, and Meta. Problems combining BFS traversal with level-based aggregation (such as sums or averages) are common follow-ups to standard zigzag level order traversal questions.

What data structure is used in Zigzag Level Sum of Binary Tree?

A queue is the primary data structure used for level order traversal in BFS. Some implementations also use a deque or stack to control traversal direction for zigzag patterns. For DFS solutions, recursion with an array or list indexed by depth is typically used.

What is the time complexity of Zigzag Level Sum of Binary Tree?

The optimal solution runs in O(n) time because every node in the binary tree is processed exactly once during traversal. Space complexity is O(w) for BFS where w is the maximum width of the tree, or O(h) for DFS where h is the tree height.

3902. Zigzag Level Sum of Binary Tree - Practice Online

You are given the root of a binary tree.

Traverse the tree level by level using a zigzag pattern:

At odd-numbered levels (1-indexed), traverse nodes from left to right.
At even-numbered levels, traverse nodes from right to left.

While traversing a level in the specified direction, process nodes in order and stop immediately before the first node that violates the condition:

At odd levels: the node does not have a left child.
At even levels: the node does not have a right child.

Only the nodes processed before this stopping condition contribute to the level sum.

Return an integer array ans where ans[i] is the sum of the node values that are processed at level i + 1.

Example 1:

Input: root = [5,2,8,1,null,9,6]

Output: [5,8,0]

Explanation:

At level 1, nodes are processed left to right. Node 5 is included, thus ans[0] = 5.
At level 2, nodes are processed right to left. Node 8 is included, but node 2 lacks a right child, so processing stops, thus ans[1] = 8.
At level 3, nodes are processed left to right. The first node 1 lacks a left child, so no nodes are included, and ans[2] = 0.
Thus, ans = [5, 8, 0].

Example 2:

Input: root = [1,2,3,4,5,null,7]

Output: [1,5,0]

Explanation:

At level 1, nodes are processed left to right. Node 1 is included, thus ans[0] = 1.
At level 2, nodes are processed right to left. Nodes 3 and 2 are included since both have right children, thus ans[1] = 3 + 2 = 5.
At level 3, nodes are processed left to right. The first node 4 lacks a left child, so no nodes are included, and ans[2] = 0.
Thus, ans = [1, 5, 0].

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
-10⁵ <= Node.val <= 10⁵