Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root. It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases. A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val. A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.   　　Example 1: 　　Input: preorder = [8,5,1,7,10,12] Output: [8,5,10,1,7,null,12]

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* bstFromPreorder(vector<int>& preorder) {
        // 就是一个简单的递归 找到分段的地方 找到左子树和右子树的分界点
        return digui(preorder,0,preorder.size()-1);
    }
        
    TreeNode * digui(vector<int>& preorder,int left,int right){
        if (left>right) return nullptr; 
        TreeNode *node=new TreeNode(preorder[left]);
        int index=left+1;
        //找到分割点
        for(;index<=right;++index){
            if(preorder[index]>preorder[left]) break; 
        }
        node->left=digui(preorder,left+1,index-1);
        node->right=digui(preorder,index,right);
        return node;
    }
};