Tree-----宽度优先遍历2(103. Binary Tree Zigzag Level Order Traversal)
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原题目
有了上一题的经验,就是把每次的宽搜循环放进去就是了,只不过在某些特定的行要进行一些操作,比如这里就是把第偶数行倒序
/** * 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>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); if (root == null) { return result; } Queue q1 = new LinkedList<>(); Queue q2 = new LinkedList<>(); q1.add(root); int counter = 0; while (q2.isEmpty() && !q1.isEmpty()) { List<Integer> item = new ArrayList<>(); counter++; while (!q1.isEmpty()) { TreeNode node = (TreeNode) q1.poll(); item.add(node.val); if (node.left != null) { q2.add(node.left); } if (node.right != null) { q2.add(node.right); } } if (counter % 2 == 0) { List<Integer> newItem = new ArrayList<>(); for (int i = item.size() - 1; i >= 0; i--) { newItem.add(item.get(i)); } result.add(newItem); } else { result.add(item); } q1 = new LinkedList<>(q2); q2.clear(); } return result; }}
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