LeetCode 144. Binary Tree Preorder Traversal (javascript)

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

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;
};