Last Updated: 25 Mar, 2021

Moderate

## Problem statement

#### 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 to true and set 'value' 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 value to  -1, and divide the current grid into four sub-grids as shown in the image.

Repeat the above steps for each of the children with the proper sub-grid.
``````

##### Input Format
``````The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.

The 1st line of each test case contains one integer 'N', representing the size of the square matrix.

The next 'N' lines contain 'N' integers separated by spaces describing rows of matrix 'matrix' (each element of ‘matrix’ is either 0 or 1).
``````
##### Output Format
``````The output represents the Quad-Tree in a level order manner, with each node represented as a comma-separated value isLeaf, value.

If the value of “isLeaf” or “value” is 1 we represent it as 1, 1, and if the value of “isLeaf” or “value” is 0 we represent it as 0, 0, and if the “value” is -1 we represent it as -1.
``````
##### Note :
``````You do not need to print anything. It has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````1 <= T <= 10
N = 2^y where 0 <= y <= 10
matrix[i][j] = 0 or 1

Time Limit: 1 second
``````

## Approaches

### 01 Approach

The idea here is to divide the grid into four equal parts the topLeft, topRight, bottomLeft, bottomRight. And then construct the Quad-Tree in a recursive manner. All four children of a node determine whether it should be a leaf node or not.

A currentNode is a leaf node only if all of its four children’s topLeft.value, topRight.value, bottomLeft.value, bottomRight.value are equal and topLeft.isLeaf, topRight.leaf, bottomLeft.leaf, bottomRight.leaf are true. Then we set current.value to any of its children’s value and set currentNode.isLeaf to True.

Otherwise, set currentNode.value to either 0 or 1 and set currentNode.isLeaf to False, and connect the four children links topLeft, topRight, bottomLeft, bottomRight to currentNode.

The algorithm is as follows:

• Create a helper function with parameters matrix, x, y, size, and return type QuadTree, where matrix represents a 2D array, (x, y) represents the starting coordinate of the 2D array, size is the current dimension of the grid, and QuadTree represents the structure of the tree itself.
• If size == 1 i.e this node is a leaf node since its size is 1.
• Create a newNode with the value same as matrix[x][y], isLeaf set to true, topLeft link as null, topRight link as null, bottomLeft link as null, bottomRight as null.
• Create a node currentNode of type QuadTree.
• Declare four variables topLeft, topRight, bottomLeft, bottomRight.
• Recurse in all four equal cuts i.e top-left, top-right, bottom-left, bottom -right of dimension size / 2.
• helper(matrix, x, y, size / 2) and set the value returned by it to topLeft, since the start of the top-left grid is (x, y).
• helper(matrix, x, y + size / 2, len / 2) and set the value returned by it to topRight, since the start of the top-right grid is (x, y + size / 2).
• helper(matrix, x + size / 2, y, size / 2) and set the value returned by it to bottomLeft, since the start of the bottom-left grid is (x + size / 2, y + size / 2).
• helper(matrix, x + size / 2, y + size / 2, size / 2) and set the value returned by it to bottomRight, since the start of the bottom-right grid is (x + size / 2, y + size / 2).
• If all four children are leaf nodes and all of their values are equal, then it’s a leaf node,
• Set currentNode.value to topLeft.value.
• Set currentNode.isLeaf to True.
• Otherwise,
• Connect all four children to currentNode.
• Set currentNode.topLeft to topLeft.
• Set currentNode.topRight to topRight.
• Set currentNode.bottomLeft to bottomLeft.
• Set currentNode.bottomRight to bottomRight.
• Set currentNode.isLeaf to False.
• Set currentNode.value to -1.
• Return the currentNode.