forked from JojoYang666/Leetcode-1-300
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_107_BinaryTreeLevelOrderTraversalII.java
79 lines (70 loc) · 2.34 KB
/
_107_BinaryTreeLevelOrderTraversalII.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
67
68
69
70
71
72
73
74
75
76
77
78
79
package leetcode_1To300;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _107_BinaryTreeLevelOrderTraversalII {
/**
* 107. Binary Tree Level Order Traversal II (102 follow up)
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
[
[15,7],
[9,20],
[3]
]
time : O(n)
space : O(n)
* @param root
* @return
*/
public static List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new LinkedList<>();
if (root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left != null) queue.offer(cur.left);
if (cur.right != null) queue.offer(cur.right);
list.add(cur.val);
}
res.add(0, list);
}
return res;
}
public static List<List<Integer>> levelOrderBottom2(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
helper(res, root, 0);
return res;
}
public static void helper(List<List<Integer>> res, TreeNode root, int level) {
if (root == null) return;
if (level >= res.size()) {
res.add(0, new ArrayList<>());
}
res.get(res.size() - level - 1).add(root.val);
helper(res, root.left, level + 1);
helper(res, root.right, level + 1);
}
}