Binary Tree Inorder Traversal - easy

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] Explanation:

Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,2,6,5,7,1,3,9,8]

Solution

/**
 * Definition for a binary tree node.
 
 *  Left -> Root -> Right
 
 * 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;
 *     }
 * }
 */
import java.util.Stack;
 
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        return traversal(root, new ArrayList<>());
    }
 
    public List<Integer> traversal(TreeNode root, List<Integer> visited) {
        TreeNode cur = root;
 
        if (cur == null) {
            return visited;
        } 
 
        traversal(cur.left, visited);
        visited.add(cur.val);
        traversal(cur.right, visited);
 
        return visited; // because visited is a reference type
 
    }
 
}

• Let recursion handle it for you!
  • The problem was that i was thinking of everything too complex when i just needed to simplify everything