Problem #102: Binary Tree Level Order Traversal

Intuition & Strategy Breakdown

Use BFS with a queue. Process nodes level by level, recording values of all nodes at current queue size into a level list.

Python Solution

from collections import deque

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        
        res = []
        q = deque([root])
        
        while q:
            level_size = len(q)
            level = []
            for _ in range(level_size):
                node = q.popleft()
                level.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            res.append(level)
            
        return res

Java Solution

class Solution {
    public List> levelOrder(TreeNode root) {
        List> res = new ArrayList<>();
        if (root == null) return res;
        
        Queue q = new LinkedList<>();
        q.offer(root);
        
        while (!q.isEmpty()) {
            int levelSize = q.size();
            List level = new ArrayList<>();
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = q.poll();
                level.add(node.val);
                if (node.left != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
            res.add(level);
        }
        return res;
    }
}