1 //[8,5,1,7,10,12]
 2 class Solution
 3 {
 4     public:
 5         TreeNode* solve(vector<int>& preorder,int l,int r)
 6         {
 7             if(l>r)
 8                 return NULL;
 9             TreeNode *root = (TreeNode*)malloc(sizeof(TreeNode));
10             root -> val = preorder[l];
11             if(l==r)
12             {
13                 root -> left = root -> right = NULL;
14                 return root;
15             }
16             int nr = -1;
17             for(int i = l+1;i <= r;i ++)
18             {
19                 if(nr==-1 && preorder[i]>root->val)
20                     nr = i;
21             }
22             if(nr==-1)
23             {
24                 root->right = NULL;
25                 root->left = solve(preorder,l+1,r);
26                 return root;
27             }
28             root -> left = solve(preorder,l+1,nr-1);
29             root -> right = solve(preorder,nr,r);
30             return root;
31         }
32         TreeNode* bstFromPreorder(vector<int>& preorder)
33         {
34             return solve(preorder,0,preorder.size()-1);
35         }
36 };

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。