# Question

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


# Solution

# 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 len(postorder) == 0:
return None
if len(postorder) == 1:
return TreeNode(postorder[0])

root = postorder[-1]
root_node = TreeNode(root)

idx = 0
while inorder[idx] != root:
idx += 1

inorder_left = inorder[:idx]
postorder_left = postorder[:idx]
left_node = self.buildTree(inorder_left, postorder_left)
root_node.left = left_node

inorder_right = inorder[idx + 1:]
postorder_right = postorder[idx:-1]
right_node = self.buildTree(inorder_right, postorder_right)
root_node.right = right_node

return root_node