LeetCode #144 — EASY

Binary Tree Preorder Traversal

Build confidence with an intuition-first walkthrough focused on stack fundamentals.

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The Problem

Problem Statement

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [1,2,4,5,6,7,3,8,9]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [1]

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?

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Stack · Tree

Example 1

[1,null,2,3]

Example 2

[1,2,3,4,5,null,8,null,null,6,7,9]

Example 3

[]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #144: Binary Tree Preorder Traversal
/**
 * 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 {
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> preorderTraversal(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        ans.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }
}

// Accepted solution for LeetCode #144: Binary Tree Preorder Traversal
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		ans = append(ans, root.Val)
		dfs(root.Left)
		dfs(root.Right)
	}
	dfs(root)
	return
}

# Accepted solution for LeetCode #144: Binary Tree Preorder Traversal
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root):
            if root is None:
                return
            ans.append(root.val)
            dfs(root.left)
            dfs(root.right)

        ans = []
        dfs(root)
        return ans

// Accepted solution for LeetCode #144: Binary Tree Preorder Traversal
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, ans: &mut Vec<i32>) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        ans.push(node.val);
        Self::dfs(&node.left, ans);
        Self::dfs(&node.right, ans);
    }

    pub fn preorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = vec![];
        Self::dfs(&root, &mut ans);
        ans
    }
}

// Accepted solution for LeetCode #144: Binary Tree Preorder Traversal
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        ans.push(root.val);
        dfs(root.left);
        dfs(root.right);
    };
    dfs(root);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.