0144 - Binary Tree Preorder Traversal

0144 - Binary Tree Preorder Traversal

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

Examples

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

Input: root = [] Output: []

Input: root = [1] Output: [1]

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

Java Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
  public List<Integer> preorderTraversal(TreeNode root) {
    LinkedList<TreeNode> stack = new LinkedList<>();
    LinkedList<Integer> result = new LinkedList<>();

    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.add(p.val);
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            p = node.right;   
        }
    }
    return result;
  }
}