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);
    }
};