105. 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.

class Solution {
public:
    TreeNode* createTree(vector<int>& preorder,int pre_start,int pre_end,vector<int>& inorder,int in_start,int in_end)
    {
        if(pre_start>pre_end)
           return NULL;
        int root = preorder[pre_start];
        int index;
        for(int i=in_start;i<=in_end;i++)
        {
            if(inorder[i] == root)
              index = i;
        }
        int len = index - in_start;
        TreeNode* left = createTree(preorder,pre_start+1,pre_start+len,inorder,in_start,index-1);
        TreeNode* right = createTree(preorder,pre_start+len+1,pre_end,inorder,index+1,in_end);
        
        TreeNode* node = new TreeNode(root);
        node->left = left;
        node->right = right;
        return node;
        
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size()==0||inorder.size()==0)
          return NULL;
        return createTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
};