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

Code:

# 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):
        if not inorder or not postorder:
            return None
        
        root = TreeNode(postorder.pop())
        inorderIndex = inorder.index(root.val)

        root.right = self.buildTree(inorder[inorderIndex+1:], postorder)
        root.left = self.buildTree(inorder[:inorderIndex], postorder)
        
        return root