text: # Time:  O(n)
  # Space: O(n)
  #
  # 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 a  binary tree node
  class TreeNode:
      def __init__(self, x):
          self.val = x
          self.left = None
          self.right = None
  
  class Solution:
      # @param inorder, a list of integers
      # @param postorder, a list of integers
      # @return a tree node
      def buildTree(self, inorder, postorder):
          lookup = {}
          for i, num in enumerate(inorder):
              lookup[num] = i
          return self.buildTreeRecu(lookup, postorder, inorder, len(postorder), 0, len(inorder))
      
      def buildTreeRecu(self, lookup, postorder, inorder, post_end, in_start, in_end):
          if in_start == in_end:
              return None
          node = TreeNode(postorder[post_end - 1])
          i = lookup[postorder[post_end - 1]]
          node.left = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1 - (in_end - i - 1), in_start, i)
          node.right = self.buildTreeRecu(lookup, postorder, inorder, post_end - 1, i + 1, in_end)
          return node
  
  if __name__ ==  "__main__":
      inorder = [2, 1, 3]
      postorder = [2, 3, 1]
      result = Solution().buildTree(inorder, postorder)
      print result.val
      print result.left.val
      print result.right.val