[Leetcode]Verify Preorder Sequence in Binary Search Tree

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree.

You may assume each number in the sequence is unique.

Follow up:

Could you do it using only constant space complexity?
void partion(vector<int>&preorder,int start,int end,int &right){
int root = preorder[start];
for(int i = start + 1;i < end;i++){
if(preorder[i] > root){
right = i;
break;
}
}
}
//[start,end)
bool verifyPreorderSub(vector<int>&preorder,int start,int end){
if((end-start) < 2)return true;
int root = preorder[start];
//divide array into two parts
int left = start + 1,right = end;
partion(preorder,start,end,right);
//check BST property
for(int i = left;i < right;i++){
if(preorder[i] > root)return false;
}
for(int i = right;i < end;i++){
if(preorder[i] < root)return false;
}
return verifyPreorderSub(preorder,left,right)&&verifyPreorderSub(preorder,right,end);
}
/*algorithm brute force
1)A[0] is root, divide A[1..n] into two parts A[1,right-1] and A[right,n]
2)check each element A[i] in left tree A[i] < root and each element A[j] in right tree, A[j] > root
3)recrusive call preorder for child array A[1..,right-1] and A[right,n]
time O(n*n) space O(1)
*/
bool verifyPreorder(vector<int>&preorder) {
return verifyPreorderSub(preorder,0,preorder.size());
}