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's
or all0's
) setisLeaf
True and setval
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 setval
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:
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:
quadTree1
andquadTree2
are both valid Quad-Trees each representing an * n
grid.n == 2x
where0 <= 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);
}
};
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#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
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