Zigzag Level Extremes — Problem Statement & Solution Guide
Problem Description
You are provided with the root of a binary tree. Your task is to analyze the tree level by level, starting from the root at depth 0 and proceeding to the deepest leaf nodes. For each distinct depth level, identify the value of the leftmost node and the value of the rightmost node as they appear in a standard left-to-right breadth-first traversal. If a level contains only a single node, that node's value is used for both the leftmost and rightmost positions. Return a list of pairs, where each pair consists of the leftmost and rightmost values for that specific level, ordered from the root level to the deepest level.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Zigzag Level Extremes"
WHY DOES IT MATTER?
Level‑order traversal is a foundational pattern for any problem that requires processing nodes in breadth‑first order, such as serialization, finding the minimum depth, or connecting next right pointers. Extracting extremes per level is a natural extension that appears in UI rendering trees, network topology analysis, and hierarchical data summarization.
OPTIMIZATION CHALLENGE
The key insight is to avoid storing the whole level; instead, track only two values while iterating through the queue. This reduces auxiliary memory from O(N) (if you kept a list per level) to O(W), the width of the tree, and keeps the runtime strictly linear.
REAL-WORLD CONNECTION
Think of a multi‑storey building where each floor has rooms arranged left to right. The leftmost and rightmost rooms on each floor are the entry and exit points for evacuation routes; identifying them quickly is analogous to our algorithm streaming floor plans without storing the entire floor layout.
During an interview, start by stating the BFS queue approach, then immediately mention that you only need the first and last node of each level, so you can compute them on the fly. This shows you understand both correctness and space efficiency.
COMPLEXITY AT A GLANCE
O(N)O(W)Core Theory — Why This Approach?
The problem reduces to a level‑order (BFS) traversal of a binary tree, where each level is processed in the order nodes appear from left to right. By capturing the first and last node encountered at each depth, we obtain the leftmost and rightmost extremes. A naive solution might attempt to store every node of a level in a list and then index the extremes, which inflates both time (due to repeated list operations) and space (because the entire level is kept). The optimal paradigm leverages a single queue to stream nodes level by level, recording the value of the node when the loop index is 0 (leftmost) and when the index equals levelSize‑1 (rightmost). This eliminates the need for auxiliary containers per level and guarantees linear work relative to the number of nodes.
Because each node is visited exactly once, the algorithm scales to trees with millions of nodes, whereas approaches that repeatedly traverse sub‑trees or rebuild level arrays suffer from O(N^2) time in degenerate cases (e.g., a skewed tree). The BFS queue naturally respects the breadth‑first order, ensuring that the leftmost node is the first dequeued at a depth and the rightmost is the last. By updating the extremes on the fly, we achieve O(N) time and O(W) auxiliary space, where W is the maximum width of the tree (worst‑case O(N) for a perfectly balanced tree).
Interview Questions on This Problem
Q1How would you modify the algorithm to also return the sum of values at each level while still reporting the leftmost and rightmost nodes?
Maintain an additional accumulator variable for each level; as you dequeue nodes, add their values to the sum. After processing the level, store a tuple (leftmost, rightmost, sum) for that depth.
Q2If the tree is extremely deep (depth > 10^5) but each level contains only one node, what changes, if any, are needed to avoid stack overflow?
Since the solution uses an iterative BFS with a queue, it already avoids recursion and thus stack overflow; no changes are required beyond ensuring the queue can handle up to O(depth) elements.
Q3Can you compute the leftmost and rightmost nodes using a depth‑first search (DFS) instead of BFS? What would be the trade‑offs?
Yes, by passing the current depth to a recursive DFS and storing the first encountered node per depth as the leftmost and overwriting the stored value each time you visit a node at that depth, the last stored value becomes the rightmost. The trade‑off is O(N) time but O(H) recursion stack space, where H is tree height, which may be problematic for very deep trees.
Examples
Input
root = [1, 2, 3, 4, 5, 6, 7]
Output
[[1, 1], [2, 3], [4, 7]]
Explanation: Level 0 contains only node 1, so the pair is [1, 1]. Level 1 contains nodes 2 and 3; the leftmost is 2 and the rightmost is 3, resulting in [2, 3]. Level 2 contains nodes 4, 5, 6, and 7; the leftmost is 4 and the rightmost is 7, resulting in [4, 7].
Input
root = [10, 20, null, 30, 40]
Output
[[10, 10], [20, 20], [30, 40]]
Explanation: Level 0 has node 10, yielding [10, 10]. Level 1 has only node 20 (since the right child is null), yielding [20, 20]. Level 2 has nodes 30 and 40; the leftmost is 30 and the rightmost is 40, yielding [30, 40].
Input
root = [5, null, 15, null, 25, 35, null]
Output
[[5, 5], [15, 15], [25, 35]]
Explanation: Level 0 is [5, 5]. Level 1 is [15, 15]. Level 2 contains nodes 25 and 35. The leftmost node at this level is 25 and the rightmost is 35, so the pair is [25, 35].
Constraints
- The number of nodes in the tree is in the range [1, 10^5].
- -10^9 <= Node.val <= 10^9.
- The tree is a valid binary tree.
- The depth of the tree will not exceed 10^5.
Optimal Approach & Strategy
Use a single queue for BFS and capture the first and last node values while iterating through each level, eliminating the need for per‑level storage.
Brute Force Approach
Traverse the tree level by level, store every node of a level in a list, then pick the first and last elements as extremes.
Verified Code Solutions
function solution(root) {
if (!root) return [];
let result = [], queue = [root];
while (queue.length > 0) {
let levelSize = queue.length, levelNodes = [];
for (let i = 0; i < levelSize; i++) {
let node = queue.shift();
levelNodes.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
result.push(levelNodes[0]);
if (levelNodes.length > 1) result.push(levelNodes[levelNodes.length - 1]);
}
return result;
}class Solution {
public:
vector<int> solution(TreeNode* root) {
if (!root) return {};
vector<int> result;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
int levelSize = q.size();
vector<int> levelNodes;
for (int i = 0; i < levelSize; i++) {
TreeNode* node = q.front(); q.pop();
levelNodes.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
result.push_back(levelNodes[0]);
if (levelNodes.size() > 1) result.push_back(levelNodes.back());
}
return result;
}
};import java.util.Queue;
import java.util.LinkedList;
public class Solution {
public int[] solution(TreeNode root) {
if (root == null) return new int[0];
List<Integer> result = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int levelSize = queue.size();
List<Integer> levelNodes = new ArrayList<>();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
levelNodes.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.add(levelNodes.get(0));
if (levelNodes.size() > 1) result.add(levelNodes.get(levelNodes.size() - 1));
}
int[] res = new int[result.size()];
for (int i = 0; i < result.size(); i++) res[i] = result.get(i);
return res;
}
}from collections import deque
def solution(root):
if not root: return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
level_nodes = []
for _ in range(level_size):
node = queue.popleft()
level_nodes.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
result.append(level_nodes[0])
if len(level_nodes) > 1: result.append(level_nodes[-1])
return resultfunction solution(root) {
if (!root) return [];
let result = [], queue = [root];
while (queue.length > 0) {
let levelSize = queue.length, levelNodes = [];
for (let i = 0; i < levelSize; i++) {
let node = queue.shift();
levelNodes.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
result.push(levelNodes[0]);
if (levelNodes.length > 1) result.push(levelNodes[levelNodes.length - 1]);
}
return result;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.