144. Binary Tree Preorder Traversal

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]
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

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

Runtime: 68 ms, faster than 95.82% of JavaScript online submissions for Binary Tree Preorder Traversal.
Memory Usage: 38.9 MB, less than 14.52% of JavaScript online submissions for Binary Tree Preorder Traversal.

/**
 * 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) {
    const result = [];
    const stack = [root];
    while(stack.length > 0) {
        node = stack.pop();
        if (node) {
            result.push(node.val);
            stack.push(node.right);
            stack.push(node.left);
        }
    }
    return result;
};

Runtime: 68 ms, faster than 95.82% of JavaScript online submissions for Binary Tree Preorder Traversal.
Memory Usage: 38.9 MB, less than 27.35% of JavaScript online submissions for Binary Tree Preorder Traversal.

/**
 * 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) {
    const result = [];
    const stack = [];
    let node = root;
    while(node || stack.length > 0) {
        if (node) {
            result.push(node.val);
            stack.push(node.right);
            node = node.left;
        } else {
            node = stack.pop();
        }
    }
    return result;
};