Total Accepted: 89586
Total Submissions: 282417
Difficulty: Easy
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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what"{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Java:
宽搜: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
29
30
31
32/**
* 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>> res = new ArrayList<List<Integer>>();
if (root == null)
return res;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
List<Integer> tmp = new ArrayList<Integer>();
int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode t = q.poll();
if (t.left != null)
q.offer(t.left);
if (t.right != null)
q.offer(t.right);
tmp.add(t.val);
}
res.add(tmp);
}
return res;
}
}
深搜:1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24/**
* 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>> res = new ArrayList<List<Integer>>();
dfs(res,root,0);
return res;
}
public void dfs(List<List<Integer>> list,TreeNode node,int deep){
if(node==null)return;
if(list.size()==deep)
list.add(new ArrayList<Integer>());
list.get(deep).add(node.val);
dfs(list, node.left, deep+1);
dfs(list, node.right, deep+1);
}
}
