用python做网站怎么赚钱/网站建设包括哪些内容

https://leetcode.com/problems/quad-tree-intersection/description/

我觉得是用意挺好的一题目。求两个四叉树的逻辑union，可惜测试用例里面居然包含对题目外因素的检查（那个id）懒得弄了。

思路其实挺简单，但是很容易忽略一个edge case，就是当所有children 的value 都一致时合并成整个leaf Node。

/*
// 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) {if (quadTree1->val == true) {return quadTree1;} else {return quadTree2;}}else if (quadTree2->isLeaf) {if (quadTree2->val == true) {return quadTree2;} else {return quadTree1;}}Node* topLeft = intersect(quadTree1->topLeft, quadTree2->topLeft);Node* topRight = intersect(quadTree1->topRight, quadTree2->topRight);Node* bottomLeft = intersect(quadTree1->bottomLeft, quadTree2->bottomLeft);Node* bottomRight = intersect(quadTree1->bottomRight, quadTree2->bottomRight);if (topLeft->isLeaf && topRight->isLeaf && bottomLeft->isLeaf && bottomRight->isLeaf) {if (topLeft->val == topRight->val == bottomLeft->val == bottomRight->val) {return new Node(topLeft->val, true, nullptr, nullptr, nullptr, nullptr);}}return new Node(0, false, topLeft, topRight, bottomLeft, bottomRight);}
};