代码随想录算法训练营第二十一天| 669 修剪二叉搜索树、108 将有序数组转换为二叉搜索树、538 把二叉搜索树转换为累加树

一、669  修剪二叉搜索树

struct TreeNode* trimBST(struct TreeNode* root, int low, int high) {
    if(root==NULL)
        return NULL;
    if(root->val<low){
        struct TreeNode* right =trimBST(root->right,low,high);
        return right;
    }
    if(root->val>high){
        struct 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;
}

二、108 将有序数组转换为二叉搜索树

struct TreeNode*Traversal(int *nums,int left,int right){
    if(left>right)
        return NULL;
    int mid = (left+right)/2;
    struct TreeNode*root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = nums[mid];
    root->left = Traversal(nums,left,mid-1);
    root->right = Traversal(nums,mid+1,right);
    return root;
}
struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
    struct TreeNode* root = Traversal(nums,0,numsSize-1);
    return root;
}

三、538 把二叉搜索树转换为累加树

//中序遍历->左中右->右中左
int pre;//记录前一个节点的数值
void Traversal(struct TreeNode*cur){
    if(cur == NULL)
        return;
    //右
    Traversal(cur->right);
    //中
    cur->val += pre;
    pre = cur->val;
    //左
    Traversal(cur->left);
}
struct TreeNode* convertBST(struct TreeNode* root) {
    pre = 0;
    Traversal(root);
    return root;
}