description： 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] ]
本題目應該使用寬度有限搜索：DFS的方法來處理更為簡單
/** * 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<Integer>> levelOrder(TreeNode root) { List<List<Integer>> list = new ArrayList<>(); if (root == null) { return list; } Queue<TreeNode> queue = new LinkedList<>(); 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); } } list.add(level); } return list; }}新聞熱點
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