题目描述
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],
方法思路
class Solution {
//Runtime: 1 ms, faster than 80.44%
//Memory Usage: 37.4 MB, less than 35.53%
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
List<Integer> l = new ArrayList<>();
l.add(root.val);
res.add(l);
while(!queue.isEmpty()){
int size = queue.size();
while(size-- > 0){
TreeNode node = queue.poll();
if(node.left != null) queue.offer(node.left);
if(node.right != null) queue.offer(node.right);
}
if(queue.size() > 0){
List<Integer> list = new ArrayList<>();
for(TreeNode node : queue)
list.add(node.val);
res.add(list);
}
}
return res;
}
}