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* visit(vector<int> &preorder, int lbegin, int lend, vector<int> &inorder, int rbegin, int rend, map<int, int> &table)
{
if (lbegin > lend)
{
return NULL;
}
TreeNode *root = new TreeNode(preorder[lbegin]);
int pos = table[preorder[lbegin]];
int leftLen = pos - rbegin;
root->left = visit(preorder, lbegin+1, lbegin+leftLen, inorder, rbegin, pos-1, table);
root->right = visit(preorder, lbegin+leftLen+1, lend, inorder, pos+1, rend, table);

return root;
}

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

map<int, int> table;
for (int i = 0; i < size; i++)
{
table[inorder[i]] = i;
}

return visit(preorder, 0, size-1, inorder, 0, size-1, table);
}
};