Leetcode: Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder 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* build(vector<int> &inorder,int left1,int right1,vector<int> &postorder,int left2,int right2)
	{
		if(right1-left1 != right2-left2)return NULL;
		if(right1>=inorder.size() || right2>=postorder.size())return NULL;

		if(left1==right1 && left2==right2)
		{
			TreeNode* root = new TreeNode(inorder[left1]);
			return root;
		}else if(left1<right1 && left2<right2)
		{
			TreeNode* root = new TreeNode(postorder[right2]);
			int i;
			for(i = right1; i >= left1; i--)
				if(inorder[i] == postorder[right2])break;
			if(i<left1)return NULL;
			root->left = build(inorder,left1,i-1,postorder,left2,right2+i-right1-1);
			root->right = build(inorder,i+1,right1,postorder,right2+i-right1,right2-1);
			return root;
		}else
			return NULL;
	}
	TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Note: The Solution object is instantiated only once.
        return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
};