-
Notifications
You must be signed in to change notification settings - Fork 2
/
LevelOrderTraversalBinaryTree.java
66 lines (55 loc) · 1.54 KB
/
LevelOrderTraversalBinaryTree.java
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
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
package Leetcode;
import java.util.*;
/**
* @author kalpak
*
* 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],
* 3
* / \
* 9 20
* / \
* 15 7
* return its level order traversal as:
* [
* [3],
* [9,20],
* [15,7]
* ]
*
*/
public class LevelOrderTraversalBinaryTree {
public static List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
Deque<TreeNode> queue = new LinkedList<>();
if(root == null)
return result;
queue.add(root);
while(!queue.isEmpty()) {
int len = queue.size();
List<Integer> temp = new ArrayList<>();
while(len > 0) {
TreeNode current = queue.poll();
temp.add(current.val);
if(current.left != null)
queue.offer(current.left);
if(current.right != null)
queue.offer(current.right);
len--;
}
result.add(temp);
}
return result;
}
public static void main(String[] args) {
TreeNode root = new TreeNode(12);
root.left = new TreeNode(7);
root.right = new TreeNode(1);
root.left.left = new TreeNode(9);
root.right.left = new TreeNode(10);
root.right.right = new TreeNode(5);
System.out.println((levelOrder(root)));
}
}