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) {}
 * };
 */
struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    TreeNode *buildTree(std::vector<int> &preorder, std::vector<int> &inorder) {
		return buildBinTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
private:
	TreeNode *buildBinTree(std::vector<int> &preorder,int preStart,int preEnd,std::vector<int> &inorder,int inStart,int inEnd)
	{
		if(preStart > preEnd || inStart > inEnd) return NULL;
		TreeNode *root = new TreeNode(preorder[preStart]);
		int rootIndex = 0;
		for (int i = inStart; i <= inEnd; i++)
		{
			if(inorder[i] == root->val)
			{
				rootIndex = i;
				break;
			}
		}
		int len = rootIndex - inStart;
		root->left = buildBinTree(preorder,preStart+1,preStart+len,inorder,inStart,rootIndex-1);
		root->right = buildBinTree(preorder,preStart+len+1,preEnd,inorder,rootIndex+1,inEnd);
		return root;
	}
};

leetcode - Construct Binary Tree from Preorder and Inorder Traversal