Logical OR of Two Binary Grids Represented as QuadTrees
This problem asks to perform a logical bitwise OR operation between two binary matrices represented by quadtrees and return a new quadtree representing the resultant matrix. Here is how we can implement a solution for this in Python.
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 class Solution:
def intersect(self, quadTree1: 'Node', quadTree2: 'Node') > 'Node':
# If either of the trees is a leaf and its value is True (1), the result is True (1)
if quadTree1.isLeaf:
return quadTree1 if quadTree1.val else quadTree2
elif quadTree2.isLeaf:
return quadTree2 if quadTree2.val else quadTree1
# If neither of the trees is a leaf, recurse on their children
topLeft = self.intersect(quadTree1.topLeft, quadTree2.topLeft)
topRight = self.intersect(quadTree1.topRight, quadTree2.topRight)
bottomLeft = self.intersect(quadTree1.bottomLeft, quadTree2.bottomLeft)
bottomRight = self.intersect(quadTree1.bottomRight, quadTree2.bottomRight)
# If all children are leaves and they have the same value, merge them into a single leaf
if topLeft.isLeaf and topRight.isLeaf and bottomLeft.isLeaf and bottomRight.isLeaf:
if topLeft.val == topRight.val == bottomLeft.val == bottomRight.val:
return Node(topLeft.val, True, None, None, None, None)
# Otherwise, create an internal node with these children
return Node(False, False, topLeft, topRight, bottomLeft, bottomRight)

In the solution, we recursively intersect the corresponding children of quadTree1
and quadTree2
. If at any point, we encounter a leaf node with value True (1), we return it directly, because any value OR’ed with 1 results in 1. If the leaf’s value is False (0), we return the other node, because any value OR’ed with 0 remains the same.
We keep doing this recursively until we have traversed all the children. If all children are leaves and have the same value, we merge them into a single leaf. If they’re not, we just create an internal node with these children.