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 main import ( "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, 0 overlapWidth = 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 := 4 bx1 := 0; by1 := -1; bx2 := 9; by2 := 2 fmt.Println(computeArea(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2)) ax1 = -2; ay1 = -2; ax2 = 2; ay2 = 2 bx1 = -2; by1 = -2; bx2 = 2; by2 = 2 fmt.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 main import "fmt" const null = -1 << 31 type TreeNode struct { Val int Left *TreeNode Right *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.Left curr.Left = curr.Right curr.Right = left if 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 := 1 for 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 } ```
