补签22day-优快云博客

自己AC过了，确实对递归理解深入了一些

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if(nums.empty()) return NULL;
        if(nums.size() == 1) {
            TreeNode* tmp = new TreeNode(nums[0]);
            return tmp;
        }
        int max = nums[0];
        int position = 0;
        for(int i = 0;i < nums.size();i++) {
            if(nums[i] > max) {
                max = nums[i];
                position = i;
            }
        }
        TreeNode* node = new TreeNode(max);
        vector<int> left(nums.begin(), nums.begin() + position);
        vector<int> right(nums.begin() + position + 1, nums.end());
        node->left = constructMaximumBinaryTree(left);
        node->right = constructMaximumBinaryTree(right);
        return node;
    }
};

题目2：617. 合并二叉树 - 力扣（LeetCode）

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if(root1 == NULL) return root2;
        if(root2 == NULL) return root1;
        TreeNode* node = new TreeNode(root1->val + root2->val);
        node->left = mergeTrees(root1->left, root2->left);
        node->right = mergeTrees(root1->right, root2->right);
        return node;
    }
};

题目3：700. 二叉搜索树中的搜索 - 力扣（LeetCode）

二叉搜索树定义：左孩子小于节点，右孩子大于节点，且左右子树也符合该规律

真的简单啊，其实整体思路没错，只是乱没想清楚

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(root == NULL || root->val == val) return root;
        TreeNode* reslut;
        if(val > root->val) reslut = searchBST(root->right, val);
        if(val < root->val) reslut = searchBST(root->left, val);
        return reslut;
    }
};

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        while(root != NULL) {
            if(val > root->val) root = root->right;
            else if(val < root->val) root = root->left;
            else {
                return root;
            }
        }
        return NULL;
    }
};

题目4：98. 验证二叉搜索树 - 力扣（LeetCode）

class Solution {
public:
    long long maxval = LONG_MIN;
    bool isValidBST(TreeNode* root) {
        if(root == NULL) return true;
        bool left = isValidBST(root->left);
        if(root->val > maxval) {
            maxval = root->val;
        }else return false;
        bool right = isValidBST(root->right);
        return left && right;
    }   
};