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],
return its level order traversal as: [ [3], [9,20], [15,7] ]node
用寬度優先搜索的方法, 用while循環遍歷樹的每一層, for循環作每一層內的遍歷code
時間 O(n) 空間O(n)it
class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) { return res; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> level = new ArrayList<>(); for (int i =0; i< size; i++) { TreeNode node = queue.poll(); level.add(node.val); if (node.left != null) { queue.offer(node.left); } if (node.right != null) { queue.offer(node.right); } } res.add(level); } return res; } }