106.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.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / 
  9  20
    /  
   15   7

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if not len(inorder) or not len(postorder):
            return None
        root = TreeNode(postorder[-1])
        pos = inorder.index(root.val)
        inorderleft,inorderright = inorder[:pos],inorder[pos+1:]
        postorderleft,postorderright = postorder[:pos],postorder[pos:pos+len(inorderright)]
        root.left = self.buildTree(inorderleft,postorderleft)
        root.right = self.buildTree(inorderright,postorderright)
        return root