leetcode-11-dfs

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

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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;
};

　　　　

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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