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Construct Quad Tree

MediumUpdated: Jul 29, 2025
Practice on:
LeetCode

Problem

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

Return the root of the Quad-Tree representing the grid.

Notice that you can assign the value of a node to True or False when isLeaf is False, and both are accepted in the answer.

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.
  • isLeaf: True if the node is leaf node on the tree or False if the node has the 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:

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.

Examples

Example 1:

\Huge \begin{array}{|c|c|} \hline \colorbox{orange} 0 & \colorbox{white} 1 \\\ \hline \colorbox{Cerulean} 1 & \colorbox{green} 0 \\\ \hline \end{array}
graph TD
    classDef orange fill:#FFA500,stroke:#333,stroke-width:2px;
    classDef white fill:#FFF,stroke:#333,stroke-width:2px;
    classDef blue fill:#00BFFF,stroke:#333,stroke-width:2px;
    classDef green fill:#32CD32,stroke:#333,stroke-width:2px;
    classDef root fill:#EEE,stroke:#333,stroke-width:2px;
    Root("Root<br>isLeaf:0<br>val:1")
    A("isLeaf:1<br>val:0")
    B("isLeaf:1<br>val:1")
    C("isLeaf:1<br>val:1")
    D("isLeaf:1<br>val:0")
    Root -- Top Left --> A
    Root -- Top Right --> B
    Root -- Bottom Left --> C
    Root -- Bottom Right --> D
    class A orange;
    class B white;
    class C blue;
    class D green;
    class Root root;

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 represnts False and 1 represents True in the photo representing the Quad-Tree.

Example 2:

\Huge \begin{array}{|c|c|c|c|c|c|c|c|} \hline \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{yellow} 0 & \colorbox{yellow} 0 & \colorbox{lightgray} 0 & \colorbox{lightgray} 0 \\\ \hline \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{yellow} 0 & \colorbox{yellow} 0 & \colorbox{lightgray} 0 & \colorbox{lightgray} 0 \\\ \hline \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{Thistle} 1 & \colorbox{Thistle} 1 & \colorbox{LimeGreen} 1 & \colorbox{LimeGreen} 1 \\\ \hline \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{orange} 1 & \colorbox{Thistle} 1 & \colorbox{Thistle} 1 & \colorbox{LimeGreen} 1 & \colorbox{LimeGreen} 1 \\\ \hline \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 \\\ \hline \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 \\\ \hline \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 \\\ \hline \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{Cerulean} 1 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 & \colorbox{lightgreen} 0 \\\ \hline \end{array}
graph TD
    classDef orange fill:#FFA500,stroke:#333,stroke-width:2px;
    classDef blue fill:#00BFFF,stroke:#333,stroke-width:2px;
    classDef green fill:#32CD32,stroke:#333,stroke-width:2px;
    classDef root fill:#EEE,stroke:#333,stroke-width:2px;
    classDef yellow fill:#FFFF99,stroke:#333,stroke-width:2px;
    classDef lightgray fill:#D3D3D3,stroke:#333,stroke-width:2px;
    classDef lightpurple fill:#E6CCFF,stroke:#333,stroke-width:2px;
    classDef lightgreen fill:#90EE90,stroke:#333,stroke-width:2px;
    classDef white fill:#FFF,stroke:#333,stroke-width:2px;
    Root("Root<br>isLeaf:0<br>val:1")
    TL("isLeaf:1<br>val:1")
    TR("isLeaf:0<br>val:1")
    BL("isLeaf:1<br>val:1")
    BR("isLeaf:1<br>val:0")
    TR_TL("isLeaf:1<br>val:0")
    TR_TR("sLeaf:1<br>val:0")
    TR_BL("isLeaf:1<br>val:1")
    TR_BR("isLeaf:1<br>val:1")
    Root -- Top Left --> TL
    Root -- Top Right --> TR
    Root -- Bottom Left --> BL
    Root -- Bottom Right --> BR
    TR -- Top Left --> TR_TL
    TR -- Top Right --> TR_TR
    TR -- Bottom Left --> TR_BL
    TR -- Bottom Right --> TR_BR
    class Root root;
    class TL orange;
    class BL blue;
    class BR green;
    class TR white;
    class TR_TL yellow;
    class TR_TR lightgray;
    class TR_BL lightpurple;
    class TR_BR lightgreen;

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:

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;
    }
}

Solution

Method 1 – Recursive Divide and Conquer (Quad Tree Construction)

Intuition

The main idea is to recursively divide the grid into four quadrants until each quadrant contains only a single value (all 0s or all 1s). At each step, we check if the current subgrid is uniform. If it is, we create a leaf node; otherwise, we continue dividing. This approach leverages the quad tree's structure, where each node represents a region of the grid, and leaf nodes represent uniform regions.

Approach

  1. Define a recursive function that takes the current subgrid boundaries (rowStart, colStart, rowEnd, colEnd).
  2. Check if all values in the current subgrid are the same:
    • If yes, return a leaf node with isLeaf = true and val set to the value of the subgrid.
    • If not, divide the subgrid into four equal parts (top left, top right, bottom left, bottom right).
  3. Recursively construct each of the four children using the corresponding subgrid boundaries.
  4. Return a non-leaf node with isLeaf = false and the four children attached.
  5. Start the recursion with the entire grid.

Complexity

  • ⏰ Time complexity: O(n^2 log n) in the worst case, as each level divides the grid and checks all cells in the region.
  • 🧺 Space complexity: O(n^2) for the tree structure and recursion stack.

Code

C++
class Node {
public:
    bool val;
    bool isLeaf;
    Node *topLeft, *topRight, *bottomLeft, *bottomRight;
    Node() : val(false), isLeaf(false), topLeft(nullptr), topRight(nullptr), bottomLeft(nullptr), bottomRight(nullptr) {}
    Node(bool v, bool l) : val(v), isLeaf(l), topLeft(nullptr), topRight(nullptr), bottomLeft(nullptr), bottomRight(nullptr) {}
    Node(bool v, bool l, Node* tl, Node* tr, Node* bl, Node* br) : val(v), isLeaf(l), topLeft(tl), topRight(tr), bottomLeft(bl), bottomRight(br) {}
};

class Solution {
public:
    Node* construct(vector<vector<int>>& grid) {
        return build(grid, 0, 0, grid.size() - 1, grid[0].size() - 1);
    }
    Node* build(vector<vector<int>>& grid, int r1, int c1, int r2, int c2) {
        bool isLeaf = true;
        int val = grid[r1][c1];
        for (int i = r1; i <= r2 && isLeaf; ++i)
            for (int j = c1; j <= c2; ++j)
                if (grid[i][j] != val) { isLeaf = false; break; }
        if (isLeaf) return new Node(val == 1, true);
        int rm = (r1 + r2) / 2, cm = (c1 + c2) / 2;
        return new Node(false, false,
            build(grid, r1, c1, rm, cm),
            build(grid, r1, cm + 1, rm, c2),
            build(grid, rm + 1, c1, r2, cm),
            build(grid, rm + 1, cm + 1, r2, c2));
    }
};
Go
type Node struct {
    Val         bool
    IsLeaf      bool
    TopLeft     *Node
    TopRight    *Node
    BottomLeft  *Node
    BottomRight *Node
}

func construct(grid [][]int) *Node {
    var build func(int, int, int, int) *Node
    build = func(r1, c1, r2, c2 int) *Node {
        val := grid[r1][c1]
        isLeaf := true
        for i := r1; i <= r2 && isLeaf; i++ {
            for j := c1; j <= c2; j++ {
                if grid[i][j] != val {
                    isLeaf = false
                    break
                }
            }
        }
        if isLeaf {
            return &Node{Val: val == 1, IsLeaf: true}
        }
        rm, cm := (r1+r2)/2, (c1+c2)/2
        return &Node{
            Val: false, IsLeaf: false,
            TopLeft: build(r1, c1, rm, cm),
            TopRight: build(r1, cm+1, rm, c2),
            BottomLeft: build(rm+1, c1, r2, cm),
            BottomRight: build(rm+1, cm+1, r2, c2),
        }
    }
    n := len(grid)
    return build(0, 0, n-1, n-1)
}
Java
class Solution {
    public Node construct(int[][] grid) {
        return build(grid, 0, 0, grid.length - 1, grid[0].length - 1);
    }
    private Node build(int[][] grid, int rowStart, int colStart, int rowEnd, int colEnd) {
        boolean isLeaf = true;
        int val = grid[rowStart][colStart];
        for (int i = rowStart; i <= rowEnd && isLeaf; i++) {
            for (int j = colStart; j <= colEnd; j++) {
                if (grid[i][j] != val) {
                    isLeaf = false;
                    break;
                }
            }
        }
        if (isLeaf) {
            return new Node(val == 1, true, null, null, null, null);
        }
        int rowMid = (rowStart + rowEnd) / 2;
        int colMid = (colStart + colEnd) / 2;
        Node topLeft = build(grid, rowStart, colStart, rowMid, colMid);
        Node topRight = build(grid, rowStart, colMid + 1, rowMid, colEnd);
        Node bottomLeft = build(grid, rowMid + 1, colStart, rowEnd, colMid);
        Node bottomRight = build(grid, rowMid + 1, colMid + 1, rowEnd, colEnd);
        return new Node(false, false, topLeft, topRight, bottomLeft, bottomRight);
    }
}
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

class Solution:
    def construct(self, grid: list[list[int]]) -> Node:
        def build(r1, c1, r2, c2):
            val = grid[r1][c1]
            isLeaf = all(grid[i][j] == val for i in range(r1, r2+1) for j in range(c1, c2+1))
            if isLeaf:
                return Node(val == 1, True)
            rm, cm = (r1 + r2) // 2, (c1 + c2) // 2
            return Node(False, False,
                build(r1, c1, rm, cm),
                build(r1, cm+1, rm, c2),
                build(rm+1, c1, r2, cm),
                build(rm+1, cm+1, r2, c2))
        n = len(grid)
        return build(0, 0, n-1, n-1)
Kotlin
class Node(
    var `val`: Boolean,
    var isLeaf: Boolean,
    var topLeft: Node? = null,
    var topRight: Node? = null,
    var bottomLeft: Node? = null,
    var bottomRight: Node? = null
)

class Solution {
    fun construct(grid: Array<IntArray>): Node? {
        fun build(r1: Int, c1: Int, r2: Int, c2: Int): Node? {
            val v = grid[r1][c1]
            var isLeaf = true
            for (i in r1..r2) for (j in c1..c2) if (grid[i][j] != v) { isLeaf = false; break }
            if (isLeaf) return Node(v == 1, true)
            val rm = (r1 + r2) / 2
            val cm = (c1 + c2) / 2
            return Node(false, false,
                build(r1, c1, rm, cm),
                build(r1, cm+1, rm, c2),
                build(rm+1, c1, r2, cm),
                build(rm+1, cm+1, r2, c2))
        }
        val n = grid.size
        return build(0, 0, n-1, n-1)
    }
}
TypeScript
class Node {
    val: boolean;
    isLeaf: boolean;
    topLeft: Node | null;
    topRight: Node | null;
    bottomLeft: Node | null;
    bottomRight: Node | null;
    constructor(val: boolean, isLeaf: boolean, topLeft?: Node | null, topRight?: Node | null, bottomLeft?: Node | null, bottomRight?: Node | null) {
        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: number[][]): Node | null {
    function build(r1: number, c1: number, r2: number, c2: number): Node {
        const val = grid[r1][c1];
        let isLeaf = true;
        for (let i = r1; i <= r2 && isLeaf; i++)
            for (let j = c1; j <= c2; j++)
                if (grid[i][j] !== val) { isLeaf = false; break; }
        if (isLeaf) return new Node(val === 1, true);
        const rm = Math.floor((r1 + r2) / 2), cm = Math.floor((c1 + c2) / 2);
        return new Node(false, false,
            build(r1, c1, rm, cm),
            build(r1, cm+1, rm, c2),
            build(rm+1, c1, r2, cm),
            build(rm+1, cm+1, r2, c2));
    }
    const n = grid.length;
    return build(0, 0, n-1, n-1);
}
Rust
pub struct Node {
    pub val: bool,
    pub is_leaf: bool,
    pub top_left: Option<Box<Node>>,
    pub top_right: Option<Box<Node>>,
    pub bottom_left: Option<Box<Node>>,
    pub bottom_right: Option<Box<Node>>,
}

impl Node {
    pub fn new(val: bool, is_leaf: bool) -> Self {
        Node { val, is_leaf, top_left: None, top_right: None, bottom_left: None, bottom_right: None }
    }
    pub fn with_children(val: bool, is_leaf: bool, tl: Node, tr: Node, bl: Node, br: Node) -> Self {
        Node {
            val,
            is_leaf,
            top_left: Some(Box::new(tl)),
            top_right: Some(Box::new(tr)),
            bottom_left: Some(Box::new(bl)),
            bottom_right: Some(Box::new(br)),
        }
    }
}

pub fn construct(grid: Vec<Vec<i32>>) -> Option<Box<Node>> {
    fn build(grid: &Vec<Vec<i32>>, r1: usize, c1: usize, r2: usize, c2: usize) -> Box<Node> {
        let val = grid[r1][c1];
        let mut is_leaf = true;
        for i in r1..=r2 {
            for j in c1..=c2 {
                if grid[i][j] != val { is_leaf = false; break; }
            }
            if !is_leaf { break; }
        }
        if is_leaf {
            return Box::new(Node::new(val == 1, true));
        }
        let rm = (r1 + r2) / 2;
        let cm = (c1 + c2) / 2;
        Box::new(Node::with_children(false, false,
            *build(grid, r1, c1, rm, cm),
            *build(grid, r1, cm+1, rm, c2),
            *build(grid, rm+1, c1, r2, cm),
            *build(grid, rm+1, cm+1, r2, c2)))
    }
    let n = grid.len();
    Some(build(&grid, 0, 0, n-1, n-1))
}

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