forked from rpj911/LeetCode_algorithm
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathGenerateTree2.java
More file actions
70 lines (59 loc) · 2.2 KB
/
GenerateTree2.java
File metadata and controls
70 lines (59 loc) · 2.2 KB
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
package Algorithms.tree;
import java.util.ArrayList;
import java.util.List;
/*
* Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
For example,
Given n = 3, your program should return all 5 unique BST's shown below.
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
* */
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; left = null; right = null; }
* }
*/
public class GenerateTree2 {
public List<TreeNode> generateTrees(int n) {
return generateTreesHelp(1, n);
}
/*
使用递归来完成，我们可以分解为2个步骤：
完成左子树，完成右子树。
如果说左子树有n种组合，右子树有m种组合，那最终的组合数就是n*m. 把这所有的组合组装起来即可
*/
public List<TreeNode> generateTreesHelp(int start, int end) {
ArrayList<TreeNode> ret = new ArrayList<TreeNode>();
// null也是一种解，也需要把它加上去。这样在组装左右子树的时候，不会出现左边没有解的情况，或
// 是右边没有解的情况
if (start > end) {
ret.add(null);
return ret;
}
for (int i = start; i <= end; i++) {
// 求出左右子树的所有的可能。
List<TreeNode> left = generateTreesHelp(start, i - 1);
List<TreeNode> right = generateTreesHelp(i + 1, end);
// 将左右子树的所有的可能性全部组装起来
for (TreeNode l: left) {
for(TreeNode r: right) {
// 先创建根节点
TreeNode root = new TreeNode(i);
root.left = l;
root.right = r;
// 将组合出来的树加到结果集合中。
ret.add(root);
}
}
}
return ret;
}
}