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 = 
Output: 
```

Example 4:

```Input: root = [1,2]
Output: [1,2]
```

Example 5:

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

Constraints:

• The number of nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`

Idea:

• Create an empty stack and push root node to stack
• Do the following while stack is not empty:
1. Pop an item from the stack and store to result array
2. Push right child of a popped item to stack
3. Push left child of a popped item to stack

Right child is pushed first: The right child is pushed before the left child to make sure that the left subtree is processed first.

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) {
let res = [];
// empty tree case:
if (root === null) return res;
let stack = [];

// preorder visit: root -> left -> right
// so the right child is pushed before the left child
// to make sure that the left subtree is processed first.
stack.push(root);
while (stack.length !== 0) {
let cur = stack.pop();
res.push(cur.val);
if (cur.right) stack.push(cur.right);
if (cur.left) stack.push(cur.left);
}
return res;
};``````