LeetCode #94 — EASY

Binary Tree Inorder Traversal

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

Solve on LeetCode

The Problem

Problem Statement

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

Example 1:

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

Output: [1,3,2]

Explanation:

Example 2:

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

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

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 inorder 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

import java.util.*;

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode cur = root;

        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            ans.add(cur.val);
            cur = cur.right;
        }
        return ans;
    }
}

func inorderTraversal(root *TreeNode) []int {
    ans := []int{}
    stack := []*TreeNode{}
    cur := root

    for cur != nil || len(stack) > 0 {
        for cur != nil {
            stack = append(stack, cur)
            cur = cur.Left
        }
        cur = stack[len(stack)-1]
        stack = stack[:len(stack)-1]
        ans = append(ans, cur.Val)
        cur = cur.Right
    }
    return ans
}

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        stack = []
        cur = root

        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            cur = stack.pop()
            ans.append(cur.val)
            cur = cur.right

        return ans

use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = Vec::new();
        let mut stack: Vec<Rc<RefCell<TreeNode>>> = Vec::new();
        let mut cur = root;

        while cur.is_some() || !stack.is_empty() {
            while let Some(node) = cur {
                cur = node.borrow().left.clone();
                stack.push(node);
            }
            let node = stack.pop().unwrap();
            ans.push(node.borrow().val);
            cur = node.borrow().right.clone();
        }

        ans
    }
}

function inorderTraversal(root: TreeNode | null): number[] {
  const ans: number[] = [];
  const stack: TreeNode[] = [];
  let cur = root;

  while (cur || stack.length) {
    while (cur) {
      stack.push(cur);
      cur = cur.left;
    }
    cur = stack.pop()!;
    ans.push(cur.val);
    cur = cur.right;
  }
  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.