558. Quad Tree Intersection（python+cpp）

题目：

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 aN * Nboolean 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.

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.

解释：
python代码：

"""
# 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):
    #这个函数用于合并之后的结点是不是叶子结点
    def isLeaf(self,node):
        topLeft=node.topLeft
        topRight=node.topRight
        bottomLeft=node.bottomLeft
        bottomRight=node.bottomRight
        if topLeft==None and topRight==None and bottomLeft==None and bottomRight==None:
            return True
        if topLeft.isLeaf and topRight.isLeaf and bottomLeft.isLeaf and bottomRight.isLeaf and (topLeft.val==topRight.val==bottomLeft.val==bottomRight.val):
            return True
        return False
    def intersect(self, quadTree1, quadTree2):
        """
        :type quadTree1: Node
        :type quadTree2: Node
        :rtype: Node
        """
        if quadTree1==None and quadTree2==None:
            return None
        if quadTree1==None or quadTree2==None:
            return quadTree1 if quadTree2==None else quadTree2
        if quadTree1.isLeaf:
            return quadTree1 if quadTree1.val else quadTree2
        if quadTree2.isLeaf:
            return quadTree2 if quadTree2.val else quadTree1
        #如果都不是叶子结点
        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)
        if self.isLeaf(quadTree1):
            quadTree1.isLeaf=True
            quadTree1.val=quadTree1.topLeft.val
        return quadTree1

c++代码：

/*
// 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 &&!quadTree2)
            return NULL;
        else if(!quadTree1 or !quadTree2)
            return quadTree1?quadTree1:quadTree2;
        if(quadTree1->isLeaf)
            return quadTree1->val?quadTree1:quadTree2;
        if(quadTree2->isLeaf)
            return quadTree2->val?quadTree2:quadTree1;
        //两个结点都不是叶子结点
        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);
        }
        //如果quadTree1合并以后变成了叶子结点。
        if (isLeaf(quadTree1))
        {
            quadTree1->isLeaf=true;
            quadTree1->val=quadTree1->topLeft->val;
        }
        return quadTree1;
            
    }
    bool isLeaf(Node* node)
    {
        Node* topLeft=node->topLeft;
        Node* topRight=node->topRight;
        Node* bottomLeft=node->bottomLeft;
        Node* bottomRight=node->bottomRight;
        if(!topLeft && !topRight && !bottomLeft && !bottomRight)
            return true;
        if(topLeft->isLeaf && topRight->isLeaf && bottomLeft->isLeaf && bottomRight->isLeaf &&(topLeft->val==topRight->val)&&(topRight->val==bottomRight->val)&&(bottomRight->val==bottomLeft->val))
            return true;
        return false;
    }
};

总结：
就是递归问题，但是合并的时候需要考虑的特殊情况比较多，这一点需要注意一下。