# CONSTRUCT BINARY TREE FROM INORDER AND POSTORDER TRAVERSAL

• ``````# 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
``````

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