Leetcode 0144. Binary Tree Preorder Traversal

Given the `root` of a binary tree, return the preorder traversal of its nodes' values.　 Example 1:　 Input: root = [1,null,2,3], Output: [1,3,2]　 Example 2:　 Input: root = [], Output: []　

Easy

Description

Given the root of a binary tree, return the preorder traversal of its nodes’ values.

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Follow up

• Recursive solution is trivial, could you do it iteratively?

Solution

/**
 * 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 {
  /**
   * Recursion
   * Time Complexity: BigO(n)
   * Space Complexity: BigO(n)
   */
  public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> list = new ArrayList<>();
    preorder(root, list);
    return list;
  }

  /**
   * Depth First Traversal: preorder
   * Node -> Left -> Right
   */
  public void preorder(TreeNode root, List<Integer> list) {
    if (root == null) return;
    list.add(root.val);
    preorder(root.left, list);
    preorder(root.right, list);
  }
}