目录
223. 矩形面积 Rectangle Area 🌟🌟
226. 翻转二叉树 Invert Binary Tree 🌟
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223. 矩形面积 Rectangle Area
给你 二维 平面上两个 由直线构成且边与坐标轴平行/垂直 的矩形，请你计算并返回两个矩形覆盖的总面积。
每个矩形由其 左下 顶点和 右上 顶点坐标表示：
- 第一个矩形由其左下顶点
(ax1, ay1)和右上顶点(ax2, ay2)定义。 - 第二个矩形由其左下顶点
(bx1, by1)和右上顶点(bx2, by2)定义。
示例 1：
输入：ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2 输出：45
示例 2：
输入：ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2 输出：16
提示：
-10^4 <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 10^4
代码：
package mainimport ("fmt"
)func computeArea(ax1 int, ay1 int, ax2 int, ay2 int, bx1 int, by1 int, bx2 int, by2 int) int {area1 := (ax2 - ax1) * (ay2 - ay1)area2 := (bx2 - bx1) * (by2 - by1)overlapWidth, overlapHeight := 0, 0overlapWidth = min(ax2, bx2) - max(ax1, bx1)overlapHeight = min(ay2, by2) - max(ay1, by1)overlapArea := max(overlapWidth, 0) * max(overlapHeight, 0)return area1 + area2 - overlapArea
}func max(a, b int) int {if a > b {return a}return b
}func min(a, b int) int {if a < b {return a}return b
}func main() {ax1 := -3; ay1 := 0; ax2 := 3; ay2 := 4bx1 := 0; by1 := -1; bx2 := 9; by2 := 2fmt.Println(computeArea(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2))ax1 = -2; ay1 = -2; ax2 = 2; ay2 = 2bx1 = -2; by1 = -2; bx2 = 2; by2 = 2fmt.Println(computeArea(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2))
}
输出：
45
16
226. 翻转二叉树 Invert Binary Tree
给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。
示例 1：
输入：root = [4,2,7,1,3,6,9] 输出：[4,7,2,9,6,3,1]
示例 2：
输入：root = [2,1,3] 输出：[2,3,1]
示例 3：
输入：root = [] 输出：[]
提示：
- 树中节点数目范围在
[0, 100]内 -100 <= Node.val <= 100
代码：
package mainimport "fmt"const null = -1 << 31type TreeNode struct {Val intLeft *TreeNodeRight *TreeNode
}func invertTree(root *TreeNode) *TreeNode {if root == nil {return nil}queue := []*TreeNode{root}for len(queue) > 0 {curr := queue[0]queue = queue[1:]left := curr.Leftcurr.Left = curr.Rightcurr.Right = leftif curr.Left != nil {queue = append(queue, curr.Left)}if curr.Right != nil {queue = append(queue, curr.Right)}}return root
}func buildTree(nums []int) *TreeNode {if len(nums) == 0 {return nil}root := &TreeNode{Val: nums[0]}Queue := []*TreeNode{root}idx := 1for idx < len(nums) {node := Queue[0]Queue = Queue[1:]if nums[idx] != null {node.Left = &TreeNode{Val: nums[idx]}Queue = append(Queue, node.Left)}idx++if idx < len(nums) && nums[idx] != null {node.Right = &TreeNode{Val: nums[idx]}Queue = append(Queue, node.Right)}idx++}return root
}func levelOrder(root *TreeNode) []int {res := []int{}if root == nil {return res}Queue := []*TreeNode{root}for len(Queue) > 0 {level := []int{}n := len(Queue)for i := 0; i < n; i++ {cur := Queue[0]Queue = Queue[1:]level = append(level, cur.Val)if cur.Left != nil {Queue = append(Queue, cur.Left)}if cur.Right != nil {Queue = append(Queue, cur.Right)}}res = append(res, level...)}return res
}func ArrayToString(arr []int) string {res := "["for i := 0; i < len(arr); i++ {res += fmt.Sprint(arr[i]) //对[]int数组可以用strconv.Itoa(arr[i])if i != len(arr)-1 {res += ","}}return res + "]"
}func main() {nums := []int{4, 2, 7, 1, 3, 6, 9}root := buildTree(nums)fmt.Println(ArrayToString(levelOrder(root)))root = invertTree(root)fmt.Println(ArrayToString(levelOrder(root)))nums = []int{2, 1, 3}root = buildTree(nums)fmt.Println(ArrayToString(levelOrder(root)))root = invertTree(root)fmt.Println(ArrayToString(levelOrder(root)))
}
输出：
[4,2,7,1,3,6,9]
[4,7,2,9,6,3,1]
[2,1,3]
[2,3,1]
递归法：
```Go
func invertTree(root *TreeNode) *TreeNode {
if root == nil {
return nil
}
left := invertTree(root.Left)
right := invertTree(root.Right)
root.Left = right
root.Right = left
return root
}
```
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