Leetcode 102. Binary Tree Level Order Traversal

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

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Analysis:

Use queue.

Time complexity: O(n)

Space complexity: O(n)

Code is below:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List> levelOrder(TreeNode root) {
        // O(n) BFS
        
        List<List> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        Queue queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            ArrayList level = new ArrayList<>();
            int size = queue.size();
            
            for (int i = 0; i < size; i++) {
                // poll the head
                TreeNode head = queue.poll();
                level.add(head.val);
                
                // add the left subnode
                if (head.left != null) {
                    queue.offer(head.left);
                }
                
                // add the right subnode
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            result.add(level);
        }
        return result;
    }
}