-
Notifications
You must be signed in to change notification settings - Fork 7
/
binaryTreePreorderTraversal.java
63 lines (48 loc) · 1.25 KB
/
binaryTreePreorderTraversal.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
// Given a binary tree, return the preorder traversal of its nodes' values.
// Example:
// Input: [1,null,2,3]
// 1
// \
// 2
// /
// 3
// Output: [1,2,3]
// Follow up: Recursive solution is trivial, could you do it iteratively?
Recursive solution
TC: O(N)
SC: O(N)
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
preorder(root,res);
return res;
}
public void preorder(TreeNode root, List<Integer> res){
if(root == null) return;
res.add(root.val);
preorder(root.left, res);
preorder(root.right, res);
}
}
Iterative solution
TC: O(N)
SC: O(N)
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if(root == null) return res;
Stack<TreeNode> s = new Stack<>();
s.push(root);
while(!s.isEmpty()){
TreeNode curr = s.pop();
res.add(curr.val);
if(curr.right!=null){
s.push(curr.right);
}
if(curr.left!=null){
s.push(curr.left);
}
}
return res;
}
}