427. Construct Quad Tree - Solution

Q: Is Construct Quad Tree easy or hard?

Construct Quad Tree is rated Medium difficulty on LeetCode. The main challenge is implementing recursive region splitting correctly and handling quadrant boundaries. Once the divide-and-conquer pattern is clear, the implementation becomes straightforward.

Q: Construct Quad Tree Python/Java solution

Python and Java implementations typically use recursion. The function checks if a region of the grid is uniform, returns a leaf node if it is, and otherwise recursively constructs four child nodes for each quadrant. The same logic translates directly between Python and Java with minor syntax differences.

Q: How to solve Construct Quad Tree in O(n)?

An O(n) approach is possible if uniformity checks are optimized with prefix sums or a similar constant-time region query technique. Instead of scanning each submatrix, the algorithm verifies whether all cells are identical using precomputed sums. Most standard solutions without this optimization run in O(n^2 log n).

Q: What is the best approach for Construct Quad Tree?

The recursive divide and conquer approach is the most common and interview-friendly solution. It checks whether a matrix region is uniform and recursively splits it into four quadrants when mixed values appear. This method naturally matches the definition of a quad tree and runs in O(n^2 log n) time with O(log n) recursion stack space.

Q: Is Construct Quad Tree asked at Google/Amazon/Meta?

Quad tree construction and spatial partitioning problems appear in interviews at companies working with graphics, maps, or spatial indexing. Variants of this problem have appeared in interviews at large tech companies including Amazon and Google because they test divide-and-conquer reasoning and tree construction.

Q: What data structure is used in Construct Quad Tree?

The core data structure is a quad tree, a hierarchical tree where each internal node has four children representing quadrants of a square region. The algorithm processes a binary matrix and builds this tree structure to compress uniform regions efficiently.

Q: What is the time complexity of Construct Quad Tree?

The typical implementation runs in O(n^2 log n) time. Each level of recursion may scan matrix cells to check if a region is uniform, and there are log n levels when repeatedly dividing the grid into quadrants. Space complexity is O(log n) due to recursion depth, or up to O(n^2) if using an iterative queue.

The problem

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 the value of the grid and set the four children to Null and stop.
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.
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 == 2^x where 0 <= x <= 6

Approach overview

Problem Overview: Given an n x n binary matrix, build a quad tree representation. Each node represents a square region. If every value in the region is the same (all 0s or all 1s), create a leaf node. Otherwise split the region into four equal quadrants and repeat the process.

Approach 1: Recursive Division (Divide and Conquer) (Time: O(n^2 log n), Space: O(log n))

This approach directly models how a quad tree works. Start with the full matrix. Check whether all values inside the current region are identical. If they are, return a leaf node with that value. If not, split the region into four equal quadrants: top-left, top-right, bottom-left, and bottom-right. Recursively construct nodes for each quadrant and attach them as children.

The key insight is that a quad tree compresses uniform regions. Large areas with identical values become a single node instead of many cells. The recursion naturally follows the structure of divide and conquer. Each recursive call works on a smaller square until the region becomes uniform or reaches size 1.

Uniformity checking requires scanning the current submatrix. In the worst case (a checkerboard-like matrix), every level scans many cells again, leading to O(n^2 log n) total time. The recursion depth is log n, which determines the auxiliary stack space.

This approach is the most common solution used in interviews and editorials. It clearly demonstrates understanding of quad tree construction and recursive decomposition of a matrix.

Approach 2: Iterative Construction Using Queue (Time: O(n^2 log n), Space: O(n^2))

An iterative version simulates the recursion with a queue. Each queue entry stores the boundaries of a matrix region along with the node that represents it. Dequeue a region, check whether all values are identical, and mark the node as a leaf if so. If the region contains mixed values, split it into four subregions and enqueue them while creating child nodes.

This approach replaces the recursive call stack with an explicit queue, which some engineers prefer when recursion depth might be large. The logic is similar: repeatedly subdivide until each node represents a uniform region.

The complexity remains O(n^2 log n) because each region may still require scanning its cells. However, space usage can grow to O(n^2) in the worst case when many subregions are waiting in the queue. The approach still relies on the same structural properties of a tree built over the original array grid.

Recommended for interviews: The recursive divide and conquer solution is what interviewers expect. It mirrors the definition of a quad tree and produces concise code. Showing the recursive approach first demonstrates strong understanding of tree construction, while discussing the iterative queue version shows you can translate recursion into explicit data structure management.

Approach 1: Recursive Division Approach

This approach involves recursively dividing the grid into four quadrants until each subdivision is either uniform (composed of the same elements) or a minimum size is reached. The base case of the recursion is when the grid section being considered is uniform, guiding us to create a leaf node. Otherwise, you recursively create non-leaf nodes and divide the grid further.

Python

class Node:
    def __init__(self, val, isLeaf, topLeft=None, topRight=None, bottomLeft=None, bottomRight=None):
        self.val = val
        self.isLeaf = isLeaf
        self.topLeft = topLeft
        self.topRight = topRight
        self.bottomLeft = bottomLeft
        self.bottomRight = bottomRight

def construct(grid):
    def is_uniform(grid, x, y, size):
        value = grid[x][y]
        for i in range(x, x + size):
            for j in range(y, y + size):
                if grid[i][j] != value:
                    return False
        return True

    def construct_quad_tree(x, y, size):
        if size == 1 or is_uniform(grid, x, y, size):
            return Node(grid[x][y] == 1, True)
        half = size // 2
        topLeft = construct_quad_tree(x, y, half)
        topRight = construct_quad_tree(x, y + half, half)
        bottomLeft = construct_quad_tree(x + half, y, half)
        bottomRight = construct_quad_tree(x + half, y + half, half)
        return Node(True, False, topLeft, topRight, bottomLeft, bottomRight)

    return construct_quad_tree(0, 0, len(grid))

Java

class Solution {
    class Node {
        public boolean val;
        public boolean isLeaf;
        public Node topLeft;
        public Node topRight;
        public Node bottomLeft;
        public Node bottomRight;

        public Node(boolean _val, boolean _isLeaf) {
            this.val = _val;
            this.isLeaf = _isLeaf;
        }

        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;
        }
    }

    public Node construct(int[][] grid) {
        return construct(grid, 0, 0, grid.length);
    }

    private Node construct(int[][] grid, int x, int y, int size) {
        if (isUniform(grid, x, y, size)) {
            return new Node(grid[x][y] == 1, true);
        }
        int half = size / 2;
        Node topLeft = construct(grid, x, y, half);
        Node topRight = construct(grid, x, y + half, half);
        Node bottomLeft = construct(grid, x + half, y, half);
        Node bottomRight = construct(grid, x + half, y + half, half);
        return new Node(true, false, topLeft, topRight, bottomLeft, bottomRight);
    }

    private boolean isUniform(int[][] grid, int x, int y, int size) {
        int val = grid[x][y];
        for (int i = x; i < x + size; i++) {
            for (int j = y; j < y + size; j++) {
                if (grid[i][j] != val) {
                    return false;
                }
            }
        }
        return true;
    }
}

This solution defines a tree node class and a function to recursively construct the Quad-Tree. The function checks if a block in the grid is uniform and creates nodes accordingly, returning the root of the constructed tree.

Complexity

Time Complexity: O(n^2 * log(n)), where n is the length of the grid. Each division checks floor(n^2) cells for uniformity.
Space Complexity: O(log(n)), due to the recursion stack depth.

Try this approach in the editor

Approach 2: Iterative Approach Using Queue

This approach involves using a queue data structure for constructing the Quad-Tree iteratively. Initially, you enqueue the entire grid with its coordinates. By dequeuing elements and checking uniformity, you either create a leaf node or enqueue its four sub-regions for further processing.

C++

#include <iostream>
#include <queue>
#include <vector>

using namespace std;

class Node {
public:
    bool val;
    bool isLeaf;
    Node* topLeft = nullptr;
    Node* topRight = nullptr;
    Node* bottomLeft = nullptr;
    Node* bottomRight = nullptr;
    Node(bool _val, bool _isLeaf) : val(_val), isLeaf(_isLeaf) {}
};

class Solution {
public:
    Node* construct(vector<vector<int>>& grid) {
        int n = grid.size();
        return construct(0, 0, n, grid);
    }

private:
    Node* construct(int x, int y, int size, vector<vector<int>>& grid) {
        if (size == 1) {
            return new Node(grid[x][y] == 1, true);
        }

        int half = size / 2;
        Node* topLeft = construct(x, y, half, grid);
        Node* topRight = construct(x, y + half, half, grid);
        Node* bottomLeft = construct(x + half, y, half, grid);
        Node* bottomRight = construct(x + half, y + half, half, grid);

        if (topLeft->isLeaf && topRight->isLeaf && bottomLeft->isLeaf && bottomRight->isLeaf &&
            topLeft->val == topRight->val && topRight->val == bottomLeft->val && bottomLeft->val == bottomRight->val) {
            return new Node(topLeft->val, true);
        }

        return new Node(true, false, topLeft, topRight, bottomLeft, bottomRight);
    }
};

JavaScript

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

function construct(grid) {
  function isUniform(x, y, size) {
    const val = grid[x][y];
    for (let i = x; i < x + size; i++) {
      for (let j = y; j < y + size; j++) {
        if (grid[i][j] !== val) return false;
      }
    }
    return true;
  }

  function buildQuadTree(x, y, size) {
    if (isUniform(x, y, size)) {
      return new Node(grid[x][y] === 1, true);
    }

    const half = size / 2;
    const topLeft = buildQuadTree(x, y, half);
    const topRight = buildQuadTree(x, y + half, half);
    const bottomLeft = buildQuadTree(x + half, y, half);
    const bottomRight = buildQuadTree(x + half, y + half, half);

    return new Node(true, false, topLeft, topRight, bottomLeft, bottomRight);
  }

  return buildQuadTree(0, 0, grid.length);
}

The C++ solution utilizes an iterative technique by gradually processing nodes via a queue. Each iteration checks for uniform grids, constructing either leaf or non-leaf nodes, and breaks down non-uniform areas for further examination.

Complexity

Time Complexity: O(n^2), as each entry in the grid must be visited to verify uniformity.
Space Complexity: O(n^2) in the worst case, since there's a possibility of storing every grid block in separate nodes.

Try this approach in the editor

Approach 3: Default Approach

/**
 * Definition for a QuadTree node.
 * type Node struct {
 *     Val bool
 *     IsLeaf bool
 *     TopLeft *Node
 *     TopRight *Node
 *     BottomLeft *Node
 *     BottomRight *Node
 * }
 */

func construct(grid [][]int) *Node {
	var dfs func(a, b, c, d int) *Node
	dfs = func(a, b, c, d int) *Node {
		zero, one := 0, 0
		for i := a; i <= c; i++ {
			for j := b; j <= d; j++ {
				if grid[i][j] == 0 {
					zero = 1
				} else {
					one = 1
				}
			}
		}
		isLeaf := zero+one == 1
		val := isLeaf && one == 1
		node := &Node{Val: val, IsLeaf: isLeaf}
		if isLeaf {
			return node
		}
		node.TopLeft = dfs(a, b, (a+c)/2, (b+d)/2)
		node.TopRight = dfs(a, (b+d)/2+1, (a+c)/2, d)
		node.BottomLeft = dfs((a+c)/2+1, b, c, (b+d)/2)
		node.BottomRight = dfs((a+c)/2+1, (b+d)/2+1, c, d)
		return node
	}
	return dfs(0, 0, len(grid)-1, len(grid[0])-1)
}

Try this approach in the editor

Complexity comparison

Approach	Complexity
Recursive Division Approach	Time Complexity: O(n^2 * log(n)), where n is the length of the grid. Each division checks floor(n^2) cells for uniformity. Space Complexity: O(log(n)), due to the recursion stack depth.
Iterative Approach Using Queue	Time Complexity: O(n^2), as each entry in the grid must be visited to verify uniformity. Space Complexity: O(n^2) in the worst case, since there's a possibility of storing every grid block in separate nodes.
Default Approach	–

Complexity analysis

Approach	Time	Space	When to Use
Recursive Division (Divide and Conquer)	O(n^2 log n)	O(log n)	Standard interview solution. Clean recursive structure that directly models quad tree construction.
Iterative Approach Using Queue	O(n^2 log n)	O(n^2)	Useful when avoiding recursion or when implementing tree construction iteratively.

Frequently asked questions

Is Construct Quad Tree easy or hard?

Construct Quad Tree is rated Medium difficulty on LeetCode. The main challenge is implementing recursive region splitting correctly and handling quadrant boundaries. Once the divide-and-conquer pattern is clear, the implementation becomes straightforward.

Construct Quad Tree Python/Java solution

Python and Java implementations typically use recursion. The function checks if a region of the grid is uniform, returns a leaf node if it is, and otherwise recursively constructs four child nodes for each quadrant. The same logic translates directly between Python and Java with minor syntax differences.

How to solve Construct Quad Tree in O(n)?

An O(n) approach is possible if uniformity checks are optimized with prefix sums or a similar constant-time region query technique. Instead of scanning each submatrix, the algorithm verifies whether all cells are identical using precomputed sums. Most standard solutions without this optimization run in O(n^2 log n).

What is the best approach for Construct Quad Tree?

The recursive divide and conquer approach is the most common and interview-friendly solution. It checks whether a matrix region is uniform and recursively splits it into four quadrants when mixed values appear. This method naturally matches the definition of a quad tree and runs in O(n^2 log n) time with O(log n) recursion stack space.

Is Construct Quad Tree asked at Google/Amazon/Meta?

Quad tree construction and spatial partitioning problems appear in interviews at companies working with graphics, maps, or spatial indexing. Variants of this problem have appeared in interviews at large tech companies including Amazon and Google because they test divide-and-conquer reasoning and tree construction.

What data structure is used in Construct Quad Tree?

The core data structure is a quad tree, a hierarchical tree where each internal node has four children representing quadrants of a square region. The algorithm processes a binary matrix and builds this tree structure to compress uniform regions efficiently.

What is the time complexity of Construct Quad Tree?

The typical implementation runs in O(n^2 log n) time. Each level of recursion may scan matrix cells to check if a region is uniform, and there are log n levels when repeatedly dividing the grid into quadrants. Space complexity is O(log n) due to recursion depth, or up to O(n^2) if using an iterative queue.

Construct Quad Tree solution and explanation

The problem

Approach overview

Approach 1: Recursive Division Approach

Approach 2: Iterative Approach Using Queue

Approach 3: Default Approach

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Construct Quad Tree solution and explanation

The problem

Approach overview

Approach 1: Recursive Division Approach

Approach 2: Iterative Approach Using Queue

Approach 3: Default Approach

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Related problems