[LeetCode] 102 Binary Tree Level Order Traversal

Description

medium

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

---

코드

/**
 * Definition for a binary tree node.
 * public 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        TreeNode rootNode = root;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int size = 0;

        if (rootNode == null) {
            return result;
        }
        
        while (queue.size() > 0) {
            List<Integer> levelResult = new ArrayList<>();
            size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode currentNode = queue.poll();
                levelResult.add(currentNode.val);
                if (currentNode.left != null) {
                    queue.offer(currentNode.left);
                }
                if (currentNode.right != null) {
                    queue.offer(currentNode.right);
                }
            }
            result.add(levelResult);
        }

        return result;
    }
}