Logical OR of Two Binary Grids Represented as Quad-Trees

1. Convert Quad-Tree to Grid: Create a helper function that recursively traverses a Quad-Tree and populates a 2D boolean grid. For a leaf node, it fills the corresponding grid area with the node's value. For an internal node, it recurses on its four children for the four sub-quadrants.
2. Get Grids: Call the conversion function for both quadTree1 and quadTree2 to obtain grid1 and grid2. This step requires knowing the grid dimension n.
3. Perform Logical OR: Create a new resultGrid of size n x n. Iterate through every cell (i, j) and compute resultGrid[i][j] = grid1[i][j] || grid2[i][j].
4. Convert Grid to Quad-Tree: Create another helper function that recursively builds a Quad-Tree from a grid. For a given grid area, it first checks if all cells have the same value. If so, it creates a leaf node. Otherwise, it creates an internal node and recursively calls itself on the four sub-quadrants.
5. Return Result: Call the grid-to-tree conversion function on resultGrid to get the final Quad-Tree.

The algorithm consists of three main parts:

1. convertToGrid(Node root, int n): A function to transform a Quad-Tree into a 2D grid. It would internally use a recursive helper fillGrid(Node node, int r, int c, int size, boolean[][] grid). If node is a leaf, it fills the size x size subgrid starting at (r, c) with node.val. If it's not a leaf, it divides the current size x size area into four size/2 x size/2 quadrants and recursively calls itself for each child node on the corresponding quadrant.
2. performOR(grid1, grid2): A function that takes two grids and returns a new grid which is the logical OR of the two. It iterates through each cell (i, j) and computes result[i][j] = grid1[i][j] || grid2[i][j].
3. buildTree(grid, r, c, size): A recursive function to build a Quad-Tree from a 2D grid. It first checks if the size x size subgrid starting at (r, c) is uniform (all 0s or all 1s). This check takes O(size^2) time. If uniform, it returns a leaf node. If not, it creates an internal node and recursively calls itself for the four sub-quadrants to build the child nodes.

The main function would orchestrate these steps. Note that this approach is generally not recommended due to its poor performance.

1// Conceptual implementation of the buildTree helper2private Node buildTree(boolean[][] grid, int r, int c, int size) {3    // Check if the subgrid is uniform4    boolean isUniform = true;5    boolean firstVal = grid[r][c];6    for (int i = r; i < r + size; i++) {7        for (int j = c; j < c + size; j++) {8            if (grid[i][j] != firstVal) {9                isUniform = false;10                break;11            }12        }13        if (!isUniform) break;14    }15 16    if (isUniform) {17        return new Node(firstVal, true, null, null, null, null);18    } else {19        int newSize = size / 2;20        Node topLeft = buildTree(grid, r, c, newSize);21        Node topRight = buildTree(grid, r, c + newSize, newSize);22        Node bottomLeft = buildTree(grid, r + newSize, c, newSize);23        Node bottomRight = buildTree(grid, r + newSize, c + newSize, newSize);24        return new Node(false, false, topLeft, topRight, bottomLeft, bottomRight);25    }26}

1. Define a recursive function intersect(node1, node2) that returns the resulting Node.
2. Base Case 1: If node1 is a leaf (node1.isLeaf == true):
   • If node1.val is true, the OR operation results in a grid of all true. Return a new leaf node with val = true.
   • If node1.val is false, the OR operation result is determined by node2. Return node2 itself.
3. Base Case 2: If node2 is a leaf, apply the symmetric logic from Base Case 1.
4. Recursive Step: If both node1 and node2 are internal nodes (not leaves):
   • Recursively call intersect for each corresponding pair of children to get the four sub-tree results: topLeft, topRight, bottomLeft, bottomRight.
   • e.g., topLeft = intersect(node1.topLeft, node2.topLeft)
5. Merge Step: After the recursive calls return, check if the four resulting children nodes can be merged. This is possible if all four are leaf nodes and all have the same val.
   • If they can be merged, return a new single leaf node with that common value.
   • Otherwise, return a new internal node with the four children nodes obtained from the recursive calls.
6. The initial call to the function will be intersect(quadTree1, quadTree2).

The core of this approach is a recursive function, intersect(node1, node2), which computes the Quad-Tree node for the logical OR of the regions represented by node1 and node2.

The recursion follows these rules:

• Pruning with Leaf Nodes: If one of the nodes, say node1, is a leaf, we can quickly determine the result. If node1.val is true (a grid of 1s), the OR result is also a grid of 1s, so we return a new leaf node with val = true. If node1.val is false (a grid of 0s), the OR result is identical to the grid represented by node2, so we can simply return node2.
• Recursive Descent: If both node1 and node2 are internal nodes, we must recurse deeper. We call intersect on each pair of corresponding children (topLeft with topLeft, topRight with topRight, etc.).
• Merging Children: After the four recursive calls for the children return, we check if the resulting four sub-trees can be merged. If all four are leaf nodes and share the same value, they represent a uniform quadrant. We can merge them by creating a single new leaf node with that common value. Otherwise, we create a new internal node and assign the four sub-trees as its children.

This process efficiently builds the result tree from the bottom up, merging quadrants whenever possible.

1class Solution {2    public Node intersect(Node quadTree1, Node quadTree2) {3        // If quadTree1 is a leaf node4        if (quadTree1.isLeaf) {5            // If its value is true (1), the OR result is a leaf with value true.6            if (quadTree1.val) {7                return new Node(true, true, null, null, null, null);8            }9            // If its value is false (0), the OR result is simply quadTree2.10            return quadTree2;11        }12 13        // If quadTree2 is a leaf node (symmetric case)14        if (quadTree2.isLeaf) {15            if (quadTree2.val) {16                return new Node(true, true, null, null, null, null);17            }18            return quadTree1;19        }20 21        // If both are internal nodes, recurse on their children22        Node tl = intersect(quadTree1.topLeft, quadTree2.topLeft);23        Node tr = intersect(quadTree1.topRight, quadTree2.topRight);24        Node bl = intersect(quadTree1.bottomLeft, quadTree2.bottomLeft);25        Node br = intersect(quadTree1.bottomRight, quadTree2.bottomRight);26 27        // Check if all children are leaves and have the same value to merge them28        if (tl.isLeaf && tr.isLeaf && bl.isLeaf && br.isLeaf &&29            tl.val == tr.val && tr.val == bl.val && bl.val == br.val) {30            // If so, merge them into a single leaf node31            return new Node(tl.val, true, null, null, null, null);32        } else {33            // Otherwise, create a new internal node with these children34            return new Node(false, false, tl, tr, bl, br);35        }36    }37}