103. Binary Tree Zigzag Level Order Traversal LeetCode Solution

In this guide, you will get 103. Binary Tree Zigzag Level Order Traversal LeetCode Solution with the best time and space complexity. The solution to Binary Tree Zigzag Level Order Traversal problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Binary Tree Zigzag Level Order Traversal solution in C++
Binary Tree Zigzag Level Order Traversal solution in Java
Binary Tree Zigzag Level Order Traversal solution in Python
Additional Resources

103. Binary Tree Zigzag Level Order Traversal LeetCode Solution image

Problem Statement of Binary Tree Zigzag Level Order Traversal

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:

The number of nodes in the tree is in the range [0, 2000].
-100 <= Node.val <= 100

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

103. Binary Tree Zigzag Level Order Traversal LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
    if (root == nullptr)
      return {};

    vector<vector<int>> ans;
    deque<TreeNode*> dq{{root}};
    bool isLeftToRight = true;

    while (!dq.empty()) {
      vector<int> currLevel;
      for (int sz = dq.size(); sz > 0; --sz)
        if (isLeftToRight) {
          TreeNode* node = dq.front();
          dq.pop_front();
          currLevel.push_back(node->val);
          if (node->left)
            dq.push_back(node->left);
          if (node->right)
            dq.push_back(node->right);
        } else {
          TreeNode* node = dq.back();
          dq.pop_back();
          currLevel.push_back(node->val);
          if (node->right)
            dq.push_front(node->right);
          if (node->left)
            dq.push_front(node->left);
        }
      ans.push_back(currLevel);
      isLeftToRight = !isLeftToRight;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

103. Binary Tree Zigzag Level Order Traversal LeetCode Solution in Java

class Solution {
  public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    if (root == null)
      return new ArrayList<>();

    List<List<Integer>> ans = new ArrayList<>();
    Deque<TreeNode> dq = new ArrayDeque<>(List.of(root));
    boolean isLeftToRight = true;

    while (!dq.isEmpty()) {
      List<Integer> currLevel = new ArrayList<>();
      for (int sz = dq.size(); sz > 0; --sz)
        if (isLeftToRight) {
          TreeNode node = dq.pollFirst();
          currLevel.add(node.val);
          if (node.left != null)
            dq.offerLast(node.left);
          if (node.right != null)
            dq.offerLast(node.right);
        } else {
          TreeNode node = dq.pollLast();
          currLevel.add(node.val);
          if (node.right != null)
            dq.offerFirst(node.right);
          if (node.left != null)
            dq.offerFirst(node.left);
        }
      ans.add(currLevel);
      isLeftToRight = !isLeftToRight;
    }

    return ans;
  }
}
// code provided by PROGIEZ

103. Binary Tree Zigzag Level Order Traversal LeetCode Solution in Python

class Solution:
  def zigzagLevelOrder(self, root: TreeNode | None) -> list[list[int]]:
    if not root:
      return []

    ans = []
    dq = collections.deque([root])
    isLeftToRight = True

    while dq:
      currLevel = []
      for _ in range(len(dq)):
        if isLeftToRight:
          node = dq.popleft()
          currLevel.append(node.val)
          if node.left:
            dq.append(node.left)
          if node.right:
            dq.append(node.right)
        else:
          node = dq.pop()
          currLevel.append(node.val)
          if node.right:
            dq.appendleft(node.right)
          if node.left:
            dq.appendleft(node.left)
      ans.append(currLevel)
      isLeftToRight = not isLeftToRight

    return ans
 # code by PROGIEZ