[leetcode]_Binary Tree Level Order Traversal I && II
题目:给定一棵二叉树,对其进行层次遍历,将遍历结果存入二维链表中。
思路:二叉树的层次遍历关键在于使用Queue。
复习:
Queue的定义。LinkedList实现了Queue接口,因此可用LinkedList来初始化一个Queue。
Queue的使用。
1、offer(),入队,当队列满掉时,返回false。
2、poll(),返回队列头部元素并删除,当队列为空时,返回null。
3、peek(),返回队列头部元素不删除,当队列为空时,返回null。
代码:
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { ArrayList<ArrayList<Integer>> level = new ArrayList<ArrayList<Integer>>(); if(root == null) return level; //在面试时,可询问考官,当root== null时,返回空链表,还是null。 Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(new TreeNode(Integer.MAX_VALUE)); //作为层与层直接的间隔符 ArrayList<Integer> oneLevel = new ArrayList<Integer>(); while(!q.isEmpty()){
TreeNode node = q.poll(); if(node.val != Integer.MAX_VALUE){ oneLevel.add(node.val);
if(node.left != null) q.offer(node.left);
if(node.right != null) q.offer(node.right); }else{
if(oneLevel.size() == 0) break; //最后一个间隔符 level.add(oneLevel);
oneLevel = new ArrayList<Integer>();
q.offer(new TreeNode(Integer.MAX_VALUE)); }
}
return level;
}
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