Pre-order + post-order to in-order

Question

Task

Given the pre-order and post-order traversals of a full binary tree, return its in-order traversal.

The traversals will be represented as two lists, both containing n distinct positive integers, each uniquely identifying a node. Your program may take these lists, and output the resulting in-order traversal, using any reasonable I/O format.

You may assume the input is valid (that is, the lists actually represent traversals of some tree).

This is code-golf, so the shortest code in bytes wins.

Definitions

A full binary tree is a finite structure of nodes, represented here by unique positive integers.

A full binary tree is either a leaf, consisting of a single node:

                                      1

Or a branch, consisting of one node with two subtrees (called the left and right subtrees), each of which is, in turn, a full binary tree:

                                      1
                                    /   \
                                  …       …

Here’s a full example of a full binary tree:

                                        6
                                      /   \
                                    3       4
                                   / \     / \
                                  1   8   5   7
                                     / \
                                    2   9

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The pre-order traversal of a full binary tree is recursively defined as follows:

• The pre-order traversal of a leaf containing a node n is the list [n].
• The pre-order traversal of a branch containing a node n and sub-trees (L, R) is the list [n] + preorder(L) + preorder(R), where + is the list concatenation operator.

For the above tree, that’s [6, 3, 1, 8, 2, 9, 4, 5, 7].

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The post-order traversal of a full binary tree is recursively defined as follows:

• The post-order traversal of a leaf containing a node n is the list [n].
• The post-order traversal of a branch containing a node n and sub-trees (L, R) is the list postorder(L) + postorder(R) + [n].

For the above tree, that’s [1, 2, 9, 8, 3, 5, 7, 4, 6].

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The in-order traversal of a full binary tree is recursively defined as follows:

• The in-order traversal of a leaf containing a node n is the list [n].
• The in-order traversal of a branch containing a node n and sub-trees (L, R) is the list inorder(L) + [n] + inorder(R).

For the above tree, that’s [1, 3, 2, 8, 9, 6, 5, 4, 7].

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In conclusion: given the pair of lists [6, 3, 1, 8, 2, 9, 4, 5, 7] (pre) and [1, 2, 9, 8, 3, 5, 7, 4, 6] (post) as input, your program should output [1, 3, 2, 8, 9, 6, 5, 4, 7].

Test cases

Each test case is in the format preorder, postorder → expected output.

[8], [8] → [8]
[3,4,5], [4,5,3] → [4,3,5]
[1,2,9,8,3], [9,8,2,3,1] → [9,2,8,1,3]
[7,8,10,11,12,2,3,4,5], [11,12,10,2,8,4,5,3,7] → [11,10,12,8,2,7,4,3,5]
[1,2,3,4,5,6,7,8,9], [5,6,4,7,3,8,2,9,1] → [5,4,6,3,7,2,8,1,9]

Answer 1

Perl, 69 66 62 56 53 bytes

Includes +1 for -p

Takes postorder followed by preorder as one line separated by space on STDIN (notice the order of pre and post). Nodes are represented as unique letters (any character that is not space or newline is OK).

inpost.pl <<< "98231 12983"

inpost.pl:

#!/usr/bin/perl -p
s%(.)(.)+\K(.)(.+)\3(\1.*)\2%$4$5$3$2%&&redo;s;.+ ;;

Using the original purely numeric format needs a lot more care to exactly identify a single number and comes in at 73 bytes

#!/usr/bin/perl -p
s%\b(\d+)(,\d+)+\K,(\d+\b)(.+)\b\3,(\1\b.*)\2\b%$4$5,$3$2%&&redo;s;.+ ;;

Use as

inpostnum.pl <<< "11,12,10,2,8,4,5,3,7 7,8,10,11,12,2,3,4,5"

Answer 2

Haskell, 84 83 bytes

(a:b:c)#z|i<-1+length(fst$span(/=b)z),h<- \f->f i(b:c)#f i z=h take++a:h drop
a#_=a

Answer 3

JavaScript (ES6), 100 bytes

f=(s,t,l=t.search(s[1]))=>s[1]?f(s.slice(1,++l+1),t.slice(0,l))+s[0]+f(s.slice(l+1),t.slice(l)):s[0]

I/O is in strings of "safe" characters (e.g. letters or digits). Alternative approach, also 100 bytes:

f=(s,t,a=0,b=0,c=s.length-1,l=t.search(s[a+1])-b)=>c?f(s,t,a+1,b,l)+s[a]+f(s,t,l+2+a,l+1,c-l-2):s[a]