【leetcode】Construct Binary Tree from Preorder and Inorder Traversal

  TreeNode* addNode(vector<int> &preorder, int& start1, vector<int>& inorder, int start2, int end2)
{
	if(start1 >= preorder.size() || end2 < start2)
	{
	    //doesn't has left branch
		--start1;
		return NULL;
	}
	//construct the root node
	TreeNode *root = new TreeNode(preorder[start1]);

	int i;//the index of current root in inorder
	for ( i = start2; i <= end2; ++i)
	{
		if(inorder[i] == preorder[start1])
			break;
	}
    //construct left branch
	root->left = addNode(preorder, ++start1, inorder, start2, i - 1);
    // construct right branch
	root->right = addNode(preorder, ++start1, inorder, i + 1, end2);
	
	return root;
}

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {

	if(preorder.size() <= 0)
		return NULL;
	int start = 0;
	
	TreeNode *root = addNode(preorder, start, inorder, 0, inorder.size() - 1);
	return root;
}

 TreeNode *createTree(vector<int>& preorder, int left1, int right1, vector<int>&inorder, int left2, int right2)
    {
        if(right1 < left1 || right2 < left2 )
        return NULL;
        TreeNode *root = new TreeNode(preorder[left1]);
        
        int i = left2;
        for(; i <= right2; ++i){
            if(inorder[i] == preorder[left1])
            break;
        }
        if(i > right2)  return NULL;
        root->left = createTree(preorder, left1 + 1, left1 + i - left2, inorder, left2, i - 1);
        root->right = createTree(preorder, left1 + i - left2 + 1, right1, inorder, i + 1, right2);
        return root;
    }
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0 || inorder.size() == 0 || preorder.size() != inorder.size())
        return NULL;
        TreeNode *root = createTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
        return root;
    }