558 Quad Tree Intersection 四叉树交集
Description:
A quadtree is a tree data in which each internal node has exactly four children: topLeft, topRight, bottomLeft and bottomRight. Quad trees are often used to partition a two-dimensional space by recursively subdividing it into four quadrants or regions.
We want to store True/False information in our quad tree. The quad tree is used to represent a N * N boolean 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 and val. 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.
Example:
For example, below are two quad trees A and B:
A:
+-------+-------+ T: true
| | | F: false
| T | T |
| | |
+-------+-------+
| | |
| F | F |
| | |
+-------+-------+
topLeft: T
topRight: T
bottomLeft: F
bottomRight: F
B:
+-------+---+---+
| | F | F |
| T +---+---+
| | T | T |
+-------+---+---+
| | |
| T | F |
| | |
+-------+-------+
topLeft: T
topRight:
topLeft: F
topRight: F
bottomLeft: T
bottomRight: T
bottomLeft: T
bottomRight: F
Your task is to implement a function that will take two quadtrees and return a quadtree that represents the logical OR (or union) of the two trees.
A: B: C (A or B):
+-------+-------+ +-------+---+---+ +-------+-------+
| | | | | F | F | | | |
| T | T | | T +---+---+ | T | T |
| | | | | T | T | | | |
+-------+-------+ +-------+---+---+ +-------+-------+
| | | | | | | | |
| F | F | | T | F | | T | F |
| | | | | | | | |
+-------+-------+ +-------+-------+ +-------+-------+
Note:
Both A and B represent grids of size N * N.
N is guaranteed to be a power of 2.
If you want to know more about the quad tree, you can refer to its wiki.
The logic OR operation is defined as this: "A or B" is true if A is true, or if B is true, or if both A and B are true.
题目描述:
四叉树是一种树数据，其中每个结点恰好有四个子结点：topLeft、topRight、bottomLeft 和 bottomRight。四叉树通常被用来划分一个二维空间，递归地将其细分为四个象限或区域。
我们希望在四叉树中存储 True/False 信息。四叉树用来表示 N * N 的布尔网格。对于每个结点, 它将被等分成四个孩子结点直到这个区域内的值都是相同的。每个节点都有另外两个布尔属性：isLeaf 和 val。当这个节点是一个叶子结点时 isLeaf 为真。val 变量储存叶子结点所代表的区域的值。
示例 :
例如，下面是两个四叉树 A 和 B：
A:
+-------+-------+ T: true
| | | F: false
| T | T |
| | |
+-------+-------+
| | |
| F | F |
| | |
+-------+-------+
topLeft: T
topRight: T
bottomLeft: F
bottomRight: F
B:
+-------+---+---+
| | F | F |
| T +---+---+
| | T | T |
+-------+---+---+
| | |
| T | F |
| | |
+-------+-------+
topLeft: T
topRight:
topLeft: F
topRight: F
bottomLeft: T
bottomRight: T
bottomLeft: T
bottomRight: F
你的任务是实现一个函数，该函数根据两个四叉树返回表示这两个四叉树的逻辑或(或并)的四叉树。
A: B: C (A or B):
+-------+-------+ +-------+---+---+ +-------+-------+
| | | | | F | F | | | |
| T | T | | T +---+---+ | T | T |
| | | | | T | T | | | |
+-------+-------+ +-------+---+---+ +-------+-------+
| | | | | | | | |
| F | F | | T | F | | T | F |
| | | | | | | | |
+-------+-------+ +-------+-------+ +-------+-------+
提示：
A 和 B 都表示大小为 N * N 的网格。
N 将确保是 2 的整次幂。
如果你想了解更多关于四叉树的知识，你可以参考这个 wiki 页面。
逻辑或的定义如下：如果 A 为 True ，或者 B 为 True ，或者 A 和 B 都为 True，则 "A 或 B" 为 True。
思路:
分 4个部分递归, 相同的叶子结点需要合并
时间复杂度O(n), 空间复杂度O(1)(不考虑递归栈)
代码:
C++:
/*
// Definition for a QuadTree node.
class Node {
public:
bool val;
bool isLeaf;
Node* topLeft;
Node* topRight;
Node* bottomLeft;
Node* bottomRight;
Node() {}
Node(bool _val, bool _isLeaf, Node* _topLeft, Node* _topRight, Node* _bottomLeft, Node* _bottomRight) {
val = _val;
isLeaf = _isLeaf;
topLeft = _topLeft;
topRight = _topRight;
bottomLeft = _bottomLeft;
bottomRight = _bottomRight;
}
};
*/
class Solution
{
public:
Node* intersect(Node* quadTree1, Node* quadTree2)
{
if (quadTree1 -> isLeaf) return quadTree1 -> val ? quadTree1 : quadTree2;
else if (quadTree2 -> isLeaf) return quadTree2 -> val ? quadTree2 : quadTree1;
else
{
quadTree1 -> topLeft = intersect(quadTree1 -> topLeft, quadTree2 -> topLeft);
quadTree1 -> topRight = intersect(quadTree1 -> topRight, quadTree2 -> topRight);
quadTree1 -> bottomLeft = intersect(quadTree1 -> bottomLeft, quadTree2 -> bottomLeft);
quadTree1 -> bottomRight = intersect(quadTree1 -> bottomRight, quadTree2 -> bottomRight);
bool val = quadTree1 -> topLeft -> val;
if (quadTree1 -> topLeft -> isLeaf and quadTree1 -> topRight -> isLeaf and quadTree1 -> bottomLeft -> isLeaf and quadTree1 -> bottomRight -> isLeaf and val == quadTree1 -> topRight -> val and val == quadTree1 -> bottomLeft -> val and val == quadTree1 -> bottomRight -> val)
{
quadTree1 -> val = val;
quadTree1 -> isLeaf = true;
}
return quadTree1;
}
}
};
Java:
/*
// Definition for a QuadTree node.
class Node {
public boolean val;
public boolean isLeaf;
public Node topLeft;
public Node topRight;
public Node bottomLeft;
public Node bottomRight;
public Node() {}
public Node(boolean _val,boolean _isLeaf,Node _topLeft,Node _topRight,Node _bottomLeft,Node _bottomRight) {
val = _val;
isLeaf = _isLeaf;
topLeft = _topLeft;
topRight = _topRight;
bottomLeft = _bottomLeft;
bottomRight = _bottomRight;
}
};
*/
class Solution {
public Node intersect(Node quadTree1, Node quadTree2) {
if (quadTree1.isLeaf) return quadTree1.val ? quadTree1 : quadTree2;
else if (quadTree2.isLeaf) return quadTree2.val ? quadTree2 : quadTree1;
else {
quadTree1.topLeft = intersect(quadTree1.topLeft, quadTree2.topLeft);
quadTree1.topRight = intersect(quadTree1.topRight, quadTree2.topRight);
quadTree1.bottomLeft = intersect(quadTree1.bottomLeft, quadTree2.bottomLeft);
quadTree1.bottomRight = intersect(quadTree1.bottomRight, quadTree2.bottomRight);
boolean val = quadTree1.topLeft.val;
if (quadTree1.topLeft.isLeaf && quadTree1.topRight.isLeaf && quadTree1.bottomLeft.isLeaf && quadTree1.bottomRight.isLeaf && val == quadTree1.topRight.val && val == quadTree1.bottomLeft.val && val == quadTree1.bottomRight.val) {
quadTree1.val = val;
quadTree1.isLeaf = true;
}
return quadTree1;
}
}
}
Python:
"""
# Definition for a QuadTree node.
class Node:
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:
def intersect(self, quadTree1: 'Node', quadTree2: 'Node') -> 'Node':
if quadTree1.isLeaf:
return quadTree1 if quadTree1.val else quadTree2
elif quadTree2.isLeaf:
return quadTree2 if quadTree2.val else quadTree1
else:
quadTree1.topLeft = self.intersect(quadTree1.topLeft, quadTree2.topLeft)
quadTree1.topRight = self.intersect(quadTree1.topRight, quadTree2.topRight)
quadTree1.bottomLeft = self.intersect(quadTree1.bottomLeft, quadTree2.bottomLeft)
quadTree1.bottomRight = self.intersect(quadTree1.bottomRight, quadTree2.bottomRight)
val = quadTree1.topLeft.val
if quadTree1.topLeft.isLeaf and quadTree1.topRight.isLeaf and quadTree1.bottomLeft.isLeaf and quadTree1.bottomRight.isLeaf and val == quadTree1.topRight.val and val == quadTree1.bottomLeft.val and val == quadTree1.bottomRight.val:
quadTree1.val = val
quadTree1.isLeaf = True
return quadTree1