[Leetcode] Construct Binary Tree from Inorder and Postorder Traversal

题目：Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

由中序和后序遍历的结构重构二叉树，且没有重复。思路跟上一篇文章差不多，只是要从后序遍历的最后开始作为根节点，然后先构造右子树再左子树。

TreeNode * aftercenTotree(vector<int> &after, vector<int> &cen, int begin, int &mid, int end)
{
	TreeNode *root=NULL;
	if(begin>end-1)  
		return root;
	int i=begin;
	for(; i<end; i++)
		if( cen[i] ==after[mid] ) break;		
	if(i==end)   return root ;				

	 root=new TreeNode(after[mid]);
	 mid--;

	root->right=aftercenTotree(after,  cen, i+1, mid, end);	
	root->left=aftercenTotree(after,  cen, begin, mid, i);
	return root;
}
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) 
{
      int mid=postorder.size()-1;
      return aftercenTotree(postorder, inorder, 0, mid, postorder.size());  
}