DFS算法:
explore(G, v) visited(v) = true
previsit(v) for each edge(v, u) in E: if not visited(u): explore(u) postvisit(v)
dfs(G) for all v in V: visited(v) = false for all v in V: if not visited(v): explore(v)
应用:
1) 判断顶点u与v之间是否存在路径
2) 判断一个无向图是否连通
112. Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
解题思路:
深度优先。使用递归的方式写。直接贴代码,简单易懂。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if (root == NULL) return false; if (root->val == sum && root->left == NULL && root->right == NULL) return true; return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); } };
类似的一道题:
129. Sum Root to Leaf Numbers
解题思路:走到叶子节点的时候加上cur值。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int sumNumbers(TreeNode* root) { if (root == NULL) return 0; sum = 0; dfs(root, 0); return sum; } void dfs(TreeNode* root, int curSum) { int cur = curSum * 10 + root->val; if (root->left == NULL && root->right == NULL) sum += cur; if (root->left != NULL) dfs(root->left, cur); if (root->right != NULL) dfs(root->right, cur); } private: int sum; };
类似的题:
257. Binary Tree Paths
解题思路:
在叶子节点时将nums转换为string放入path中。注意,因为每个节点的值都压栈了,所以每个压入nums的值,最后都要出栈,不然打印路径时会有重复。
数字转字符串:sprintf
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<string> binaryTreePaths(TreeNode* root) { if (root == NULL) return path; vector<int> nums; dfs(root,nums); return path; } void dfs(TreeNode* root, vector<int> nums) { if (root->left == NULL && root->right == NULL) { nums.push_back(root->val); string str=""; for (int i = 0; i < nums.size(); i++) { if (i != 0) str += "->"; char arr[10]; sprintf(arr, "%d", nums[i]); str += arr; } path.push_back(str); nums.pop_back(); } if (root->left != NULL) { nums.push_back(root->val); dfs(root->left, nums); nums.pop_back(); } if (root->right != NULL) { nums.push_back(root->val); dfs(root->right, nums); nums.pop_back(); } } private: vector<string> path; };
拓扑排序:
1) Find a source, output it, and delete it from the graph. Repeat until the graph is empty.
2) dfs and sort with post[u] in descending order: TOPOLOGICAL-SORT(G)
1 call DFS(G) to compute finishing times post[v] for each vertex v
2 as each vertex is finished, insert it onto the front of a linked list
3 return the linked list of vertices