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 a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, int preorderStart,int preorderEnd,vector<int>& inorder,int inorderStart,int inorderEnd) {
        if(preorderEnd-preorderStart==1){
            return new TreeNode(preorder[preorderStart]);
        }
        if(preorderEnd-preorderStart==0){
            return 0;
        }
        int val = preorder[preorderStart];
        int i=inorderStart;
        for(;i<inorderEnd;i++){
            if(inorder[i] == val){
                break;
            }
        }
        TreeNode *root = new TreeNode(val);
        int leftLen = i-inorderStart;
        int rightLen = inorderEnd - i - 1;
        root->left = buildTree(preorder,preorderStart+1,preorderStart+1+leftLen,inorder,inorderStart,inorderStart+leftLen);
        root->right = buildTree(preorder,preorderStart+1+leftLen,preorderEnd,inorder,inorderStart+leftLen+1,inorderEnd);
        return root;
        
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int preorederSize = preorder.size();
        int inorederSize = inorder.size();
        if(preorederSize != inorederSize){
            return NULL;
        }
        return buildTree(preorder,0,preorder.size(),inorder,0,inorder.size());
    }
};