Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Move from brute-force thinking to an efficient approach using tree strategy.
Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[3],[20,9],[15,7]]
Example 2:
Input: root = [1] Output: [[1]]
Example 3:
Input: root = [] Output: []
Constraints:
[0, 2000].-100 <= Node.val <= 100Problem summary: Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to right, then right to left for the next level and alternate between).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[3,9,20,null,null,15,7]
[1]
[]
binary-tree-level-order-traversal)zigzag-grid-traversal-with-skip)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #103: Binary Tree Zigzag Level Order Traversal
/**
* 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 {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
boolean left = true;
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.poll();
t.add(node.val);
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
if (!left) {
Collections.reverse(t);
}
ans.add(t);
left = !left;
}
return ans;
}
}
// Accepted solution for LeetCode #103: Binary Tree Zigzag Level Order Traversal
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func zigzagLevelOrder(root *TreeNode) (ans [][]int) {
if root == nil {
return
}
q := []*TreeNode{root}
left := true
for len(q) > 0 {
t := []int{}
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
t = append(t, node.Val)
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
if !left {
for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
t[i], t[j] = t[j], t[i]
}
}
ans = append(ans, t)
left = !left
}
return
}
# Accepted solution for LeetCode #103: Binary Tree Zigzag Level Order Traversal
# 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 zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
ans = []
if root is None:
return ans
q = deque([root])
ans = []
left = 1
while q:
t = []
for _ in range(len(q)):
node = q.popleft()
t.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(t if left else t[::-1])
left ^= 1
return ans
// Accepted solution for LeetCode #103: Binary Tree Zigzag Level Order Traversal
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
pub fn zigzag_level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut ans = Vec::new();
let mut left = true;
if let Some(root_node) = root {
let mut q = VecDeque::new();
q.push_back(root_node);
while !q.is_empty() {
let mut t = Vec::new();
for _ in 0..q.len() {
if let Some(node) = q.pop_front() {
let node_ref = node.borrow();
t.push(node_ref.val);
if let Some(ref left) = node_ref.left {
q.push_back(Rc::clone(left));
}
if let Some(ref right) = node_ref.right {
q.push_back(Rc::clone(right));
}
}
}
if !left {
t.reverse();
}
ans.push(t);
left = !left;
}
}
ans
}
}
// Accepted solution for LeetCode #103: Binary Tree Zigzag Level Order Traversal
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function zigzagLevelOrder(root: TreeNode | null): number[][] {
const ans: number[][] = [];
if (!root) {
return ans;
}
const q: TreeNode[] = [root];
let left: number = 1;
while (q.length) {
const t: number[] = [];
const qq: TreeNode[] = [];
for (const { val, left, right } of q) {
t.push(val);
left && qq.push(left);
right && qq.push(right);
}
ans.push(left ? t : t.reverse());
q.splice(0, q.length, ...qq);
left ^= 1;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.