LeetCode #427 — MEDIUM

Construct Quad Tree

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

Given a n * n matrix grid of 0's and 1's only. We want to represent grid with a Quad-Tree.

Return the root of the Quad-Tree representing grid.

A Quad-Tree is a tree data structure in which each internal node has exactly four children. Besides, each node has two attributes:

  • val: True if the node represents a grid of 1's or False if the node represents a grid of 0's. Notice that you can assign the val to True or False when isLeaf is False, and both are accepted in the answer.
  • isLeaf: True if the node is a leaf node on the tree or False if the node has four children.
class Node {
    public boolean val;
    public boolean isLeaf;
    public Node topLeft;
    public Node topRight;
    public Node bottomLeft;
    public Node bottomRight;
}

We can construct a Quad-Tree from a two-dimensional area using the following steps:

  1. If the current grid has the same value (i.e all 1's or all 0's) set isLeaf True and set val to the value of the grid and set the four children to Null and stop.
  2. If the current grid has different values, set isLeaf to False and set val to any value and divide the current grid into four sub-grids as shown in the photo.
  3. Recurse for each of the children with the proper sub-grid.

If you want to know more about the Quad-Tree, you can refer to the wiki.

Quad-Tree format:

You don't need to read this section for solving the problem. This is only if you want to understand the output format here. The output represents the serialized format of a Quad-Tree using level order traversal, where null signifies a path terminator where no node exists below.

It is very similar to the serialization of the binary tree. The only difference is that the node is represented as a list [isLeaf, val].

If the value of isLeaf or val is True we represent it as 1 in the list [isLeaf, val] and if the value of isLeaf or val is False we represent it as 0.

Example 1:

Input: grid = [[0,1],[1,0]]
Output: [[0,1],[1,0],[1,1],[1,1],[1,0]]
Explanation: The explanation of this example is shown below:
Notice that 0 represents False and 1 represents True in the photo representing the Quad-Tree.

Example 2:

Input: grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
Output: [[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
Explanation: All values in the grid are not the same. We divide the grid into four sub-grids.
The topLeft, bottomLeft and bottomRight each has the same value.
The topRight have different values so we divide it into 4 sub-grids where each has the same value.
Explanation is shown in the photo below:

Constraints:

  • n == grid.length == grid[i].length
  • n == 2x where 0 <= x <= 6
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given a n * n matrix grid of 0's and 1's only. We want to represent grid with a Quad-Tree. Return the root of the Quad-Tree representing grid. A Quad-Tree is a tree data structure in which each internal node has exactly four children. Besides, each node has two attributes: val: True if the node represents a grid of 1's or False if the node represents a grid of 0's. Notice that you can assign the val to True or False when isLeaf is False, and both are accepted in the answer. isLeaf: True if the node is a leaf node on the tree or False if the node has four children. class Node { public boolean val; public boolean isLeaf; public Node topLeft; public Node topRight; public Node bottomLeft; public Node bottomRight; } We can construct a Quad-Tree from a two-dimensional area using the following steps: If the current grid has the same value (i.e all 1's or all 0's) set isLeaf True and set val to

Baseline thinking

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

Pattern signal: Array · Tree

Example 1

[[0,1],[1,0]]

Example 2

[[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
Step 02

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

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #427: Construct Quad Tree
/*
// Definition for a QuadTree node.
class Node {
    public boolean val;
    public boolean isLeaf;
    public Node topLeft;
    public Node topRight;
    public Node bottomLeft;
    public Node bottomRight;


    public Node() {
        this.val = false;
        this.isLeaf = false;
        this.topLeft = null;
        this.topRight = null;
        this.bottomLeft = null;
        this.bottomRight = null;
    }

    public Node(boolean val, boolean isLeaf) {
        this.val = val;
        this.isLeaf = isLeaf;
        this.topLeft = null;
        this.topRight = null;
        this.bottomLeft = null;
        this.bottomRight = null;
    }

    public Node(boolean val, boolean isLeaf, Node topLeft, Node topRight, Node bottomLeft, Node
bottomRight) { this.val = val; this.isLeaf = isLeaf; this.topLeft = topLeft; this.topRight =
topRight; this.bottomLeft = bottomLeft; this.bottomRight = bottomRight;
    }
};
*/

class Solution {
    public Node construct(int[][] grid) {
        return dfs(0, 0, grid.length - 1, grid[0].length - 1, grid);
    }

    private Node dfs(int a, int b, int c, int d, int[][] grid) {
        int zero = 0, one = 0;
        for (int i = a; i <= c; ++i) {
            for (int j = b; j <= d; ++j) {
                if (grid[i][j] == 0) {
                    zero = 1;
                } else {
                    one = 1;
                }
            }
        }
        boolean isLeaf = zero + one == 1;
        boolean val = isLeaf && one == 1;
        Node node = new Node(val, isLeaf);
        if (isLeaf) {
            return node;
        }
        node.topLeft = dfs(a, b, (a + c) / 2, (b + d) / 2, grid);
        node.topRight = dfs(a, (b + d) / 2 + 1, (a + c) / 2, d, grid);
        node.bottomLeft = dfs((a + c) / 2 + 1, b, c, (b + d) / 2, grid);
        node.bottomRight = dfs((a + c) / 2 + 1, (b + d) / 2 + 1, c, d, grid);
        return node;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.

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