construct-binary-tree-from-inorder-and-postorder-traversal.py

# 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