[LeetCode] 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?

Thought process:
Preorder traversal visits a tree in the order of root - left subtree - right subtree. 

Solution 1 (Recursive):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> preorder = new ArrayList<>();
        
        preorderTraversal(root, preorder);
        
        return preorder;
    }
    
    private void preorderTraversal(TreeNode root, List<Integer> preorder) {
        if (root == null) {
            return;
        }
        
        preorder.add(root.val);
        preorderTraversal(root.left, preorder);
        preorderTraversal(root.right, preorder);
    }
}

Time complexity: O(n).

The recursive solution is trivial. To solve this iteratively, we need to use a stack to simulate the function call stack.

Remember to check for null nodes to prevent null pointer exception.

Solution 2 (Iterative):
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> preorder = new ArrayList<>();
        
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (node == null) {
                continue;
            }
            
            preorder.add(node.val);
            
            stack.push(node.right);
            stack.push(node.left);
        }
        
        return preorder;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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