558. Quad Tree Intersection

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。

1/*
2// Definition for a QuadTree node.
3class Node {
4public:
5    bool val;
6    bool isLeaf;
7    Node* topLeft;
8    Node* topRight;
9    Node* bottomLeft;
10    Node* bottomRight;
11
12    Node() {}
13
14    Node(bool _val, bool _isLeaf, Node* _topLeft, Node* _topRight, Node* _bottomLeft, Node* _bottomRight) {
15        val = _val;
16        isLeaf = _isLeaf;
17        topLeft = _topLeft;
18        topRight = _topRight;
19        bottomLeft = _bottomLeft;
20        bottomRight = _bottomRight;
21    }
22};
23*/
24class Solution {
25public:
26    Node* intersect(Node* quadTree1, Node* quadTree2) {
27        if (quadTree1->isLeaf) {
28            if (quadTree1->val == true) {
29                return quadTree1;
30            } else {
31                return quadTree2;
32            }
33        }
34        else if (quadTree2->isLeaf) {
35            if (quadTree2->val == true) {
36                return quadTree2;
37            } else {
38                return quadTree1;
39            }
40        }
41        
42        Node* topLeft = intersect(quadTree1->topLeft, quadTree2->topLeft);
43        Node* topRight = intersect(quadTree1->topRight, quadTree2->topRight);
44        Node* bottomLeft = intersect(quadTree1->bottomLeft, quadTree2->bottomLeft);
45        Node* bottomRight = intersect(quadTree1->bottomRight, quadTree2->bottomRight);
46
47        if (topLeft->isLeaf && topRight->isLeaf && bottomLeft->isLeaf && bottomRight->isLeaf) {
48            if (topLeft->val == topRight->val == bottomLeft->val == bottomRight->val) {
49                return new Node(topLeft->val, true, nullptr, nullptr, nullptr, nullptr);
50            }
51        }
52        return new Node(0, false, topLeft, topRight, bottomLeft, bottomRight);
53    }
54};