刷爆leetcode Day14 DFS

6. 计算布尔二叉树的值（medium）

https://leetcode.cn/problems/evaluate-boolean-binary-tree/submissions/577469929/

/**
 * 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 {
public:
    bool evaluateTree(TreeNode* root) {
        if(root->left==nullptr)
        return root->val==1?true:false;
        bool left=evaluateTree(root->left);
        bool right=evaluateTree(root->right);
        return root->val==2?left||right:left&&right;
    }
};

7. 求根节点到叶节点数字之和（medium）

https://leetcode.cn/problems/sum-root-to-leaf-numbers/submissions/577475267/

/**
 * 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 {
public:
    int sumNumbers(TreeNode* root) {
        return _sumNumbers(root,0);
    }
    int _sumNumbers(TreeNode* root,int Presum)
    {
        Presum=Presum*10+root->val;
        if(root->left==nullptr&&root->right==nullptr)
        return Presum;
        int ret=0;
        if(root->left)
        ret+=_sumNumbers(root->left,Presum);
        if(root->right)
        ret+=_sumNumbers(root->right,Presum);
        return ret;
    }
};

8. 二叉树剪枝（medium）

https://leetcode.cn/problems/binary-tree-pruning/submissions/577479516/

/**
 * 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 {
public:
    TreeNode* pruneTree(TreeNode* root) {
        if(root==nullptr)
        return nullptr;
        root->left=pruneTree(root->left);
        root->right=pruneTree(root->right);
        if(root->left==nullptr&&root->right==nullptr&&root->val==0)
        root=nullptr;

        return root;
    }
};

9. 验证二叉搜索树（medium）

https://leetcode.cn/problems/validate-binary-search-tree/submissions/577484480/

/**
 * 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 {
    long prev=LONG_MIN;
public:

    bool isValidBST(TreeNode* root) {
        if(root==nullptr)
        return true;
        bool left=isValidBST(root->left);
        bool cur=false;
        if(root->val>prev)
        {
            cur=true;
            prev=root->val;
        }
        bool right=isValidBST(root->right);
        return left&&right&&cur;
    }
};

10. 二叉搜索树中第k小的元素（medium）

https://leetcode.cn/problems/kth-smallest-element-in-a-bst/submissions/577486024/

/**
 * 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 {
    int count;
    int ret;
public:
    int kthSmallest(TreeNode* root, int k) {
        count=k;
        _kthSmallest(root);
        return ret;
    }
    void _kthSmallest(TreeNode* root)
    {
        if(root==nullptr)return;
        _kthSmallest(root->left);
        count--;
        if(count==0)
        ret=root->val;
        _kthSmallest(root->right);
    }
};