144 - Binary Tree Preorder Traversal

144. Binary Tree Preorder Traversal

Problem Statement:

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]

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:

Java

List<Integer> list = new ArrayList<Integer>();

// Iterative Approach
public List<Integer> preorderTraversal(TreeNode root) {
  if (root == null)
    return list;
  Stack<TreeNode> stack = new Stack<>();
  stack.push(root);
  while (!stack.isEmpty()) {
    TreeNode current = stack.pop();
    list.add(current.val);
    if (current.right != null)
      stack.push(current.right);
    if (current.left != null)
      stack.push(current.left);
  }
  return list;
}

// Recursive Approach
public List<Integer> preorderTraversalRecursive(TreeNode root) {
  if (root != null) {
    list.add(root.val);
    preorderTraversalRecursive(root.left);
    preorderTraversalRecursive(root.right);
  }
  return list;
}

...