DAY11 - DFS应用

昨天写了DFS模板，今天就搭配模板放几题DFS的例题!!

void dfs(){
    if(越界或不合理状态)
        return
    for(对当前节点扩展){
        if(next_node合理&&未被访问){
            vis[next_node] = 1
            dfs(next_node)
            vis[next_node] = 0
        }
    }
}

例题实战

230. 二叉搜索树中第K小的元素
二元树&DFS
从程序码就会发现其实在应用的时候程序还是跟模板很像的，只要确认好调用的时机就好~~

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    stack<int> sta;
public:
    int kthSmallest(TreeNode* root, int k) {
        dfs(root, k);
        return sta.top();
    }
    void dfs(TreeNode* root, int k){
        if(root==nullptr){
            return;
        }
        dfs(root->left, k);
        if(sta.size()==k){
            return;
        }
        sta.push(root->val);
        dfs(root->right, k);
    }
};

剑指 Offer 36. 二叉搜索树与双向链表
这题是二元树的中序遍历
((其实就是一边中序遍历，一边指向正确的节点))

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;

    Node() {}

    Node(int _val) {
        val = _val;
        left = NULL;
        right = NULL;
    }

    Node(int _val, Node* _left, Node* _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
    Node* front = nullptr;
    Node* head = nullptr;
public:
    /*二元树的中序遍历符合双向链的顺序*/
    Node* treeToDoublyList(Node* root) {
        if(root==nullptr){
            return nullptr;
        }
        dfs(root);
        front->right = head;
        head->left = front;
        return head;
    }
    void dfs(Node* curr){
        if(curr==nullptr){
            return;
        }
        dfs(curr->left);
        if(front==nullptr){ //遍历到第一个节点
            head = curr;
        }
        else{
            front->right = curr;
        }
        curr->left = front;
        front = curr;
        dfs(curr->right);
    }
};

47. 全排列 II
排列在昨天的文章有写了一题，这边再补一题允许重复的~~

class Solution {
    vector<vector<int>> res;
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        generate(nums, 0);
        return res;
        
    }
    void generate(vector<int> nums, int n){
        if(n >= nums.size())
        {
            res.push_back(nums);
            return;
        }
        for(int i = n; i < nums.size(); i++)
        {
            if( i != n && nums[n] == nums[i])
                continue;
            swap(nums[n], nums[i]);
            generate(nums, n+1);
        }
        return;
    }

};