199. 二叉树的右视图 Binarytree Right Side View

给定一个二叉树的 根节点 root，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例 1:

输入: [1,2,3,null,5,null,4]
输出: [1,3,4]

示例 2:

输入: [1,null,3]
输出: [1,3]

示例 3:

输入: []
输出: []

提示:

二叉树的节点个数的范围是 [0,100]
-100 <= Node.val <= 100

代码：

package mainimport "fmt"const null = -1 << 31type TreeNode struct {Val   intLeft  *TreeNodeRight *TreeNode
}func (root *TreeNode) RightSideView() []int {var res []intif root == nil {return res}queue := []*TreeNode{root}for len(queue) > 0 {n := len(queue)for i := 0; i < n; i++ {node := queue[i]if i == n-1 {res = append(res, node.Val)}if node.Left != nil {queue = append(queue, node.Left)}if node.Right != nil {queue = append(queue, node.Right)}}queue = queue[n:]}return res
}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 ArrayToString(arr []int) string {res := "["for i := 0; i < len(arr); i++ {res += fmt.Sprint(arr[i])if i != len(arr)-1 {res += ","}}return res + "]"
}func main() {nums := []int{1, 2, 3, null, 5, null, 4}root := buildTree(nums)fmt.Println(ArrayToString(root.RightSideView()))nums = []int{1, null, 3}root = buildTree(nums)fmt.Println(ArrayToString(root.RightSideView()))
}

输出：

[1,3,4]
[1,3]

200. 岛屿数量 Number-of-islands

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]
输出：1

示例 2：

输入：grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

代码：

package mainimport "fmt"func numIslands(grid [][]byte) int {// 迭代整个网格，遇到陆地时就进行 DFS 搜索，找到一个区域后计数器加一m, n := len(grid), len(grid[0])var count intfor i := 0; i < m; i++ {for j := 0; j < n; j++ {if grid[i][j] == '1' {count++dfs(grid, i, j, m, n)}}}return count
}// DFS 搜索函数，将所有和当前岛屿相邻的陆地都标记为已遍历
func dfs(grid [][]byte, i, j, m, n int) {// 避免越界if i < 0 || j < 0 || i >= m || j >= n || grid[i][j] == '0' {return}// 标记当前位置为已遍历grid[i][j] = '0'// DFS 搜索四周的陆地dfs(grid, i+1, j, m, n)dfs(grid, i-1, j, m, n)dfs(grid, i, j+1, m, n)dfs(grid, i, j-1, m, n)
}func main() {grid1 := [][]byte{{'1', '1', '1', '1', '0'},{'1', '1', '0', '1', '0'},{'1', '1', '0', '0', '0'},{'0', '0', '0', '0', '0'},}fmt.Println(numIslands(grid1))grid2 := [][]byte{{'1', '1', '0', '0', '0'},{'1', '1', '0', '0', '0'},{'0', '0', '1', '0', '0'},{'0', '0', '0', '1', '1'},}fmt.Println(numIslands(grid2))
}

输出：

1
3