Binary Tree Preorder Traversal 🌲

https://leetcode.com/problems/binary-tree-preorder-traversal

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

Preorder Traversal :

Note that node 75 doesn't have left child and node 29 doesn't have right child.

The preorder traversal for above binary tree will be [70, 67, 35, 21, 20, 75, 30, 43, 29, 50, 24, 60, 65]

Implementation : Recursive

import java.util.ArrayList;
import java.util.List;

public class App {

     public static void main(String[] args) {
		TreeNode root = new TreeNode(70);
		root.left = new TreeNode(67); root.right = new TreeNode(43);
		
		root.left.left = new TreeNode(35); root.left.right = new TreeNode(75); 
		
		root.left.left.left = new TreeNode(21); root.left.left.right = new TreeNode(20);
		
		root.left.right.right = new TreeNode(30);
		
		root.right.left = new TreeNode(29); root.right.left.left = new TreeNode(50);
		
		root.right.right = new TreeNode(24); 
		
		root.right.right.left = new TreeNode(60); root.right.right.right = new TreeNode(65);
		System.out.println(inorderTraversal(root));
     }
	
	
     // Definition for a binary tree node.
     static public class TreeNode {
	      int val;
	      TreeNode left;
	      TreeNode right;
	      TreeNode(int x) { val = x; }
     }
	 
	
     public static List <Integer> inorderTraversal(TreeNode root) {
        List <Integer> res = new ArrayList<Integer>();
        helper(root, res);
        return res;
     }

     public static void helper(TreeNode root, List <Integer> res) {
        if (root != null) {
             res.add(root.val);
             helper(root.left, res);
             helper(root.right, res);
        }
     }
}

Implementation : Iterative (Stacks are awesome ...)

public List<Integer> preorderTraversal(TreeNode root) {
       List<Integer> res = new ArrayList<>(); 
       if (root == null) 
            return res;

       Stack<TreeNode> stack = new Stack<>();
       stack.push(root);

       while (!stack.empty()) {
            TreeNode curr = stack.pop();
            res.add(curr.val);
	    
            if (curr.right != null) {
                stack.push(curr.right);
            }
           
            if (curr.left != null) {
                stack.push(curr.left);
            }
      }
    return res;
}

References :