day21 二叉树8

669. 修剪二叉搜索树

递归法：

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == nullptr) {
            return root;
        } 
        if (root->val < low) {
            TreeNode* right = trimBST(root->right, low, high);
            return right;
        }
        if (root->val > high) {
            TreeNode* left = trimBST(root->left, low, high);
            return left;
        }
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);
        return root;
    }
};

迭代法：

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == nullptr) {
            return root;
        } 
        while (root && (root->val < low || root->val > high)) {
            if (root->val < low) {
                root = root->right;
            }
            else {
                root = root->left;
            }
        }
        TreeNode* cur = root;
        while (cur) {
            while (cur->left && cur->left->val < low) {
                cur->left = cur->left->right;
            }
            cur = cur->left;
        }
        cur = root;
        while (cur) {
            while (cur->right && cur->right->val > high) {
                cur->right = cur->right->left;
            }
            cur = cur->right;
        }
        return root;
    }
};

108.将有序数组转换为二叉搜索树

class Solution {
public:
    TreeNode* toBst(vector<int> &nums, int left, int right) {
        if (left >= right) {
            return nullptr;
        }
        int mid = left + (right - left) / 2;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = toBst(nums, left, mid);
        node->right = toBst(nums, mid + 1, right);
        return node;
    }

    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return toBst(nums, 0, nums.size());
    }
};

迭代法太麻烦了，不写了

538.把二叉搜索树转换为累加树

class Solution {
public:
    int sumNum = 0;
    void getSum(TreeNode* root) {
        if (root == nullptr) {
            return;
        }
        sumNum += root->val;
        getSum(root->left);
        getSum(root->right);
    }
    TreeNode* convert(TreeNode* root) {
        if (root == nullptr) {
            return root;
        }
        root->left = convert(root->left);
        sumNum -= root->val;
        root->val += sumNum;
        root->right = convert(root->right);
        return root;
    }
    
    TreeNode* convertBST(TreeNode* root) {
        getSum(root);
        return convert(root);
    }
};

二叉搜索树的左中右是从小到大的，右中左就是从大到小的，反着写更方便，不用求和了。

class Solution {
public:
    int sumNum = 0;
    TreeNode* convertBST(TreeNode* root) {
        if (root == nullptr) {
            return root;
        }
        root->right = convertBST(root->right);
        sumNum += root->val;
        root->val = sumNum;
        root->left = convertBST(root->left);
        return root;
    }
};