forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_144.java
67 lines (58 loc) · 1.57 KB
/
_144.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
package com.fishercoder.solutions;
import com.fishercoder.common.classes.TreeNode;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;
/**
* 144. Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?*/
public class _144 {
public static class Solution1 {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList();
if (root == null) {
return list;
}
Deque<TreeNode> stack = new ArrayDeque<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode curr = stack.pop();
list.add(curr.val);
/**We push right nodes onto the stack first, since they'll be popped out later than
* the left nodes, to meet the preorder: root -> left -> right. */
if (curr.right != null) {
stack.push(curr.right);
}
if (curr.left != null) {
stack.push(curr.left);
}
}
return list;
}
}
public static class Solution2 {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList();
return pre(root, list);
}
List<Integer> pre(TreeNode root, List<Integer> list) {
if (root == null) {
return list;
}
list.add(root.val);
pre(root.left, list);
pre(root.right, list);
return list;
}
}
}