Binary Tree Zigzag Level Order Traversal

🧩 Problem Statement

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]]

🔍 Approaches

1. BFS with Toggle Flag ($O(N)$)

• Concept: Standard BFS, but control the order of adding elements to the current level list.
• Algorithm:
• Use a queue for BFS.
• Maintain a boolean leftToRight = true.
• For each level:
• If leftToRight is true, add nodes to the list normally (append).
• If false, add nodes to the front of the list (or reverse the list after collecting).
• Toggle leftToRight after processing the level.
• Data Structure: A Deque for the current level list makes adding to front $O(1)$. Alternatively, just reverse() the vector at the end of the level.

⏳ Time & Space Complexity

• Time Complexity: $O(N)$.
• Space Complexity: $O(N)$ (queue width).

🚀 Code Implementations

C++

#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
if (!root) return result;
queue<TreeNode*> q;
q.push(root);
bool leftToRight = true;
while (!q.empty()) {
int size = q.size();
vector<int> level(size);
for (int i = 0; i < size; i++) {
TreeNode* node = q.front();
q.pop();
int index = leftToRight ? i : (size - 1 - i);
level[index] = node->val;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
leftToRight = !leftToRight;
result.push_back(level);
}
return result;
}
};

Python

from typing import List, Optional
from collections import deque
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]]:
if not root:
return []
result = []
queue = deque([root])
left_to_right = True
while queue:
level_size = len(queue)
current_level = deque()
for _ in range(level_size):
node = queue.popleft()
if left_to_right:
current_level.append(node.val)
else:
current_level.appendleft(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(list(current_level))
left_to_right = not left_to_right
return result

Java

import java.util.*;
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>> result = new ArrayList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
boolean leftToRight = true;
while (!queue.isEmpty()) {
int size = queue.size();
LinkedList<Integer> level = new LinkedList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (leftToRight) {
level.addLast(node.val);
} else {
level.addFirst(node.val);
}
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
leftToRight = !leftToRight;
result.add(level);
}
return result;
}
}

🌍 Real-World Analogy

Snake Printing:

Like an old dot-matrix printer or a snake game moving across the screen. Left-to-right, drop down, then right-to-left.

🎯 Summary

✅ Direction Flag: A simple boolean flag toggles the insertion order (append vs prepend) at each level.