LeetCode_Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode * constructSubtree(vector<int> &preOrder, int preStart, int preEnd, vector<int> &inOrder, int inStart, int inEnd){

// if(preStart < preEnd) return NULL ;

int temp = preOrder[preStart] ;
TreeNode *root = new TreeNode(temp) ;
if(preStart == preEnd)
return root ;

int i;
for(i = inStart ; i<= inEnd ;i++)
if(inOrder[i] == temp ) break;

root->left = preStart+1 <= preStart+i-inStart ? constructSubtree(preOrder, preStart+1, preStart+i-inStart, inOrder, inStart, i-1) : NULL;
root->right = preStart+i-inStart+1 <=  preEnd ? constructSubtree(preOrder, preStart+i-inStart+1,  preEnd, inOrder, i+1, inEnd) : NULL;

return root ;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

if(preorder.size() == 0 ) return NULL;
TreeNode *root = new TreeNode(preorder[0]) ;
if(preorder.size() == 1 ) return root ;

int i;
int temp = preorder[0] ;

for(i = 0; i< inorder.size(); i++)
if(inorder[i] == temp) break ;

root->left = 1 <= i ? constructSubtree(preorder, 1,i,inorder, 0, i-1) : NULL  ;

root->right = i+1 <= preorder.size()-1 ? constructSubtree(preorder, i+1,preorder.size()-1, inorder, i+1, inorder.size() -1): NULL ;

return root;
}

};