Algorithm
Problem Name: 558. Logical OR of Two Binary Grids Represented as Quad-Trees
A Binary Matrix is a matrix in which all the elements are either 0 or 1.
Given quadTree1 and quadTree2. quadTree1 represents a n * n binary matrix and quadTree2 represents another n * n binary matrix.
Return a Quad-Tree representing the n * n binary matrix which is the result of logical bitwise OR of the two binary matrixes represented by quadTree1 and quadTree2.
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:
- If the current grid has the same value (i.e all 1'sor all0's) setisLeafTrue and setvalto the value of the grid and set the four children to Null and stop.
- If the current grid has different values, set isLeafto False and setvalto 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:
The input/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: quadTree1 = [[0,1],[1,1],[1,1],[1,0],[1,0]] , quadTree2 = [[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]] Output: [[0,0],[1,1],[1,1],[1,1],[1,0]] Explanation: quadTree1 and quadTree2 are shown above. You can see the binary matrix which is represented by each Quad-Tree. If we apply logical bitwise OR on the two binary matrices we get the binary matrix below which is represented by the result Quad-Tree. Notice that the binary matrices shown are only for illustration, you don't have to construct the binary matrix to get the result tree.
Example 2:
Input: quadTree1 = [[1,0]], quadTree2 = [[1,0]] Output: [[1,0]] Explanation: Each tree represents a binary matrix of size 1*1. Each matrix contains only zero. The resulting matrix is of size 1*1 with also zero.
Constraints:
- quadTree1and- quadTree2are both valid Quad-Trees each representing a- n * ngrid.
- n == 2xwhere- 0 <= x <= 9.
Code Examples
#1 Code Example with Javascript Programming
Code -
                                                        Javascript Programming
const intersect = function (quadTree1, quadTree2) {
  if (quadTree1.isLeaf) {
    return quadTree1.val ? quadTree1 : quadTree2;
  }
  if (quadTree2.isLeaf) {
    return quadTree2.val ? quadTree2 : quadTree1;
  }
  const topLeft = intersect(quadTree1.topLeft, quadTree2.topLeft);
  const topRight = intersect(quadTree1.topRight, quadTree2.topRight);
  const bottomLeft = intersect(quadTree1.bottomLeft, quadTree2.bottomLeft);
  const bottomRight = intersect(quadTree1.bottomRight, quadTree2.bottomRight);
  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, null, null, null, null);
  } else {
    return new Node(false, false, topLeft, topRight, bottomLeft, bottomRight);
  }
};
Input
Output
#2 Code Example with Python Programming
Code -
                                                        Python Programming
class Solution:
    def intersect(self, q1, q2):
        if q1.isLeaf:
            return q1.val and q1 or q2
        elif q2.isLeaf:
            return q2.val and q2 or q1
        else:
            tLeft = self.intersect(q1.topLeft, q2.topLeft)
            tRight = self.intersect(q1.topRight, q2.topRight)
            bLeft = self.intersect(q1.bottomLeft, q2.bottomLeft)
            bRight = self.intersect(q1.bottomRight, q2.bottomRight)
            if tLeft.isLeaf and tRight.isLeaf and bLeft.isLeaf and bRight.isLeaf and tLeft.val == tRight.val == bLeft.val == bRight.val:
                node = Node(tLeft.val, True, None, None, None, None) 
            else:
                node = Node(False, False, tLeft, tRight, bLeft, bRight)
        return node
Input
Output
