102. Binary Tree Level Order Traversal (BFS)

BFS Traversal

力扣

問題點

所以需要利用當前 queue 的 size 當分層的數量

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

                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            result.add(subList);
        }
        return result;
    }
}