Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,2,3]
Explanation:
Example 2:
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [1,2,4,5,6,7,3,8,9]
Explanation:
Example 3:
Input: root = []
Output: []
Example 4:
Input: root = [1]
Output: [1]
Constraints:
The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
My Solution
To do the preorder traversal, we first add the current node’s value to the list, then recursively call the left subtree and then finally recursively call the right subtree.
Time complexity would be O(n), where n is the number of nodes in the tree. Space complexity would be O(1) because we only require constant extra space.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> out = new ArrayList<>();
preorder(root, out);
return out;
}
public void preorder(TreeNode root, List<Integer> list) {
if (root == null) {
return;
}
list.add(root.val);
preorder(root.left, list);
preorder(root.right, list);
}
}