Leetcode Binary Tree Level Order Traversal II Java Solution

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

Solution:
We can solve it multiple ways using Breadth First Search and Depth First Search the below solution uses Depth First Search with Back Tracking.

class HackerHeap {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
         List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (root == null) {
            return result;
        }
        dfs(root, 0, result);
        return result;
    }
    public void dfs(TreeNode root, int level, List<List<Integer>> rst) {
        if (root == null) {
            return;
        }
        if (level >= rst.size()) {
            rst.add(0, new ArrayList<Integer>());
        }
        dfs(root.left, level + 1, rst);
        dfs(root.right, level + 1, rst);
        rst.get(rst.size() - level - 1).add(root.val);
    }
}

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rajendra