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

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

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function (root) {
  //Return an empty array if root is null
  if (root === null) {
    return [];
  }
  //Create a result array
  let result = [];
  //Create a stack
  let stack = [];
  //Push root to stack
  stack.push(root);
  //WHILE stack is not empty
  while (stack.length !== 0) {
    //Set current to stack.pop()
    let current = stack.pop();
    //Push current.val to result
    result.push(current.val);
    //IF current.right is not null
    if (current.right !== null) {
      //Push current.right to stack
      stack.push(current.right);
    }
    //IF current.left is not null
    if (current.left !== null) {
      //Push current.left to stack
      stack.push(current.left);
    }
  }
  //Return result
  return result;
};