-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathBinaryTreeBuild.java
More file actions
68 lines (62 loc) · 2.5 KB
/
Copy pathBinaryTreeBuild.java
File metadata and controls
68 lines (62 loc) · 2.5 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
package offer07;
import sun.reflect.generics.tree.Tree;
import java.util.List;
import java.util.ArrayList;
public class BinaryTreeBuild {
public static class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int v){
val = v;
}
}
//@大威版
// public TreeNode buildTree(int[] preorder, int[] inorder) {
// if(preorder == null || inorder == null) return null;
// if(preorder.length == 0 || inorder.length == 0) return null;
// if(preorder.length != inorder.length) return null;
// TreeNode root = new TreeNode(preorder[0]);
// for(int i = 0; i < inorder.length; i++){
// if(preorder[0] == inorder[i]){
// root.left = buildTree(Arrays.copyOfRange(preorder,1,i + 1),
// Arrays.copyOfRange(inorder,0,i));
// root.right = buildTree(Arrays.copyOfRange(preorder,i + 1,preorder.length),
// Arrays.copyOfRange(inorder,i + 1,inorder.length));
// }
// }
// return root;
// }
public TreeNode buildTree(int[] preorder, int[] inorder) {
/*
* @Des: 根据给定的前序和中序序列，构造出对应的二叉树
* @Para:
* preorder：前序序列
* inorder: 中序序列
* @Return:
* 二叉树
**/
ArrayList<Integer> pre = new ArrayList<>();
ArrayList<Integer> in = new ArrayList<>();
for(int value: preorder){ //复制到ArrayList里，为了切片与indexof
pre.add(value);
}
for(int value: inorder){
in.add(value);
}
return build(pre.subList(0, pre.size()), in.subList(0, in.size()));//这里切片是为了转化类型
// 之所以使用Arraylist,是想使用其切片功能与根据值返回下标功能，但是查看源码后发现其返回下标
// 功能就是遍历对比，这样子的话，还不如自己遍历，然后切片使用CopyofRange()
}
private TreeNode build(List<Integer> pre, List<Integer> in){
if(pre.size() < 1){
return null;
}
TreeNode head = new TreeNode(pre.get(0));
int root = in.indexOf(pre.get(0)); //当前子树的根节点位置
//根节点
head.left = build( pre.subList(1, root+1), in.subList(0, root));
head.right = build(pre.subList(root+1, pre.size()), in.subList(root+1,in.size()));
return head;
}
}