Leetcode 0094. Binary Tree Inorder Traversal

Given the `root` of a binary tree, return the inorder 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 inorder 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> inorderTraversal(TreeNode root) {
    List<Integer> nums = new ArrayList<>();
    inorder(root, nums);
    return nums;
  }

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