Leetcode-1008 Construct Binary Search Tree from Preorder Traversal(先序遍历构造二叉树)

//[8,5,1,7,10,12]
class Solution
{
    public:
        TreeNode* solve(vector<int>& preorder,int l,int r)
        {
            if(l>r)
                return NULL;
            TreeNode *root = (TreeNode*)malloc(sizeof(TreeNode));
            root -> val = preorder[l];
            if(l==r)
            {
                root -> left = root -> right = NULL;
                return root;
            }
            int nr = -1;
            for(int i = l+1;i <= r;i ++)
            {
                if(nr==-1 && preorder[i]>root->val)
                    nr = i;
            }
            if(nr==-1)
            {
                root->right = NULL;
                root->left = solve(preorder,l+1,r);
                return root;
            }
            root -> left = solve(preorder,l+1,nr-1);
            root -> right = solve(preorder,nr,r);
            return root;
        }
        TreeNode* bstFromPreorder(vector<int>& preorder)
        {
            return solve(preorder,0,preorder.size()-1);
        }
};

这题应该叫先序遍历构建二叉搜索树。第一个肯定是根节点，往后找第一个比根节点大的数，那个就是右子树的根节点。根节点下一个就是左子树根节点。左子树根节点到右子树根节点之间就是左子树，右子树根节点到最后就是右子树。