102. Binary Tree Level Order Traversal LeetCode Solution

In this guide, you will get 102. Binary Tree Level Order Traversal LeetCode Solution with the best time and space complexity. The solution to Binary Tree 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.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Binary Tree Level Order Traversal solution in C++
4. Binary Tree Level Order Traversal solution in Java
5. Binary Tree Level Order Traversal solution in Python
6. Additional Resources

Problem Statement of Binary Tree Level Order Traversal

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[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].
-1000 <= Node.val <= 1000

Complexity Analysis

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

102. Binary Tree Level Order Traversal LeetCode Solution in C++

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

    vector<vector<int>> ans;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      vector<int> currLevel;
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode* node = q.front();
        q.pop();
        currLevel.push_back(node->val);
        if (node->left)
          q.push(node->left);
        if (node->right)
          q.push(node->right);
      }
      ans.push_back(currLevel);
    }

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

102. Binary Tree Level Order Traversal LeetCode Solution in Java

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

    List<List<Integer>> ans = new ArrayList<>();
    Queue<TreeNode> q = new ArrayDeque<>(List.of(root));

    while (!q.isEmpty()) {
      List<Integer> currLevel = new ArrayList<>();
      for (int sz = q.size(); sz > 0; --sz) {
        TreeNode node = q.poll();
        currLevel.add(node.val);
        if (node.left != null)
          q.offer(node.left);
        if (node.right != null)
          q.offer(node.right);
      }
      ans.add(currLevel);
    }

    return ans;
  }
}
// code provided by PROGIEZ

102. Binary Tree Level Order Traversal LeetCode Solution in Python

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

    ans = []
    q = collections.deque([root])

    while q:
      currLevel = []
      for _ in range(len(q)):
        node = q.popleft()
        currLevel.append(node.val)
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)
      ans.append(currLevel)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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