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:
1
2
3
4
5
|
[
[3],
[9,20],
[15,7]
]
|
解法
借助队列可以实现按层遍历二叉树，对于每一层的处理，在处理每一层前记录队列的大小即可确定本层需要处理的节点的个数，这样不会让节点的插入影响处理过程。
具体代码如下：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
|
class {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
int level = 1;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
List<Integer> tmp = new ArrayList<>();
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
TreeNode tmpnode = q.poll();
tmp.add(tmpnode.val);
if (tmpnode.left != null) {
q.offer(tmpnode.left);
}
if (tmpnode.right != null) {
q.offer(tmpnode.right);
}
}
res.add(tmp);
tmp = new ArrayList<>();
}
return res;
}
}
|
近期评论