We want to use quad trees to store an
N x N boolean grid. Each cell in the grid can only be true or false. The root node represents the whole grid. For each node, it will be subdivided into four children nodes until the values in the region it represents are all the same.
Each node has another two boolean attributes :
isLeaf is true if and only if the node is a leaf node. The
val attribute for a leaf node contains the value of the region it represents.
Your task is to use a quad tree to represent a given grid. The following example may help you understand the problem better:
8 x 8 grid below, we want to construct the corresponding quad tree:
It can be divided according to the definition above:
The corresponding quad tree should be as following, where each node is represented as a
(isLeaf, val) pair.
For the non-leaf nodes,
val can be arbitrary, so it is represented as
Nis less than
1000and guaranteened to be a power of 2.
- If you want to know more about the quad tree, you can refer to its wiki.
""" # Definition for a QuadTree node. class Node(object): def __init__(self, val, isLeaf, topLeft, topRight, bottomLeft, bottomRight): self.val = val self.isLeaf = isLeaf self.topLeft = topLeft self.topRight = topRight self.bottomLeft = bottomLeft self.bottomRight = bottomRight """ class Solution(object): def construct(self, grid): """ :type grid: List[List[int]] :rtype: Node """ def construct2(g, i1, j1, n): s = g[i1][j1] is_same = True for i in xrange(i1, i1+n): for j in xrange(j1, j1+n): if g[i][j] != s: is_same = False break if is_same: return Node(s, True, None, None, None, None) else: root = Node("*", False, None, None, None, None) root.topLeft = construct2(g, i1, j1, n/2) root.topRight = construct2(g, i1, j1+n/2, n/2) root.bottomLeft = construct2(g, i1+n/2, j1, n/2) root.bottomRight = construct2(g, i1+n/2, j1+n/2, n/2) return root return construct2(grid, 0, 0, len(grid))
class Solution: def construct(self, grid): def dfs(x, y, l): if l == 1: node = Node(grid[x][y] == 1, True, None, None, None, None) else: tLeft = dfs(x, y, l // 2) tRight = dfs(x, y + l // 2, l // 2) bLeft = dfs(x + l // 2, y, l// 2) bRight = dfs(x + l // 2, y + l // 2, l // 2) value = tLeft.val or tRight.val or bLeft.val or bRight.val if tLeft.isLeaf and tRight.isLeaf and bLeft.isLeaf and bRight.isLeaf and tLeft.val == tRight.val == bLeft.val == bRight.val: node = Node(value, True, None, None, None, None) else: node = Node(value, False, tLeft, tRight, bLeft, bRight) return node return grid and dfs(0, 0, len(grid)) or None