1 DFS和BFS

深度优先遍历一般采用回溯算法进行解决。回溯算法，其实就是dfs的过程。

void dfs(参数) {处理节点dfs(图，选择的节点); // 递归回溯，撤销处理结果
}

广度优先搜索理解为层次遍历。

797. 所有可能的路径

直接DFS就好了，有向无环图甚至不用设置一个visited数组来判断某个点是否访问过。

class Solution:def __init__(self):self.result_list = []def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:path = [0]visited = [0] * len(graph)self.traverse(0, len(graph), graph, visited, path)return self.result_listdef traverse(self, i, n, graph, visited, path):if i == n-1:self.result_list.append(path.copy())  # 注意要copyreturnfor j in graph[i]:if visited[j] == 0:path.append(j)self.traverse(j, n, graph, visited, path)  # dfspath.pop()    # 回溯return  

200. 岛屿数量

class Solution:def dfs(self, i, j, visited, grid, m, n):visited[i][j] = 1nbrs = [(i, j-1), (i-1, j), (i+1, j), (i, j+1)]for nbr in nbrs:if 0 <= nbr[0] < m and 0 <= nbr[1] < n and visited[nbr[0]][nbr[1]] == 0 and grid[nbr[0]][nbr[1]] == '1':self.dfs(nbr[0], nbr[1], visited, grid, m, n)def numIslands(self, grid: List[List[str]]) -> int:# 每次搜索到一个连通分量后重新从下一个点开始f即可total_num = 0m, n = len(grid), len(grid[0])visited = [[0 for _ in range(n)] for _ in range(m)]for i in range(m):for j in range(n):if visited[i][j] == 0 and grid[i][j] == '1':total_num += 1self.dfs(i,j,visited,grid,m,n)return total_num