# Preorder from Inorder and Postorder traversals

Given Inorder and Postorder traversals of a binary tree, print Preorder traversal.

Example:

```Input: Postorder traversal post[] = {4, 5, 2, 6, 3, 1}
Inorder traversal in[] = {4, 2, 5, 1, 3, 6}
Output: Preorder traversal 1, 2, 4, 5, 3, 6

Trversals in the above example represents following tree
1
/    \
2       3
/   \      \
4     5      6
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A naive method is to first construct the tree from given postorder and inorder, then use simple recursive method to print preorder traversal of the constructed tree. We can print preorder traversal without constructing the tree. The idea is, root is always the first item in preorder traversal and it must be the last item in postorder traversal. We first push right subtree to a stack, then left subtree and finally we push root. Finally we print contents of stack. To find boundaries of left and right subtrees in post[] and in[], we search root in in[], all elements before root in in[] are elements of left subtree and all elements after root are elements of right subtree. In post[], all elements after index of root in in[] are elements of right subtree. And elements before index (including the element at index and excluding the first element) are elements of left subtree.

## Java

 `// Java program to print Postorder traversal from given ` `// Inorder and Preorder traversals. ` `import` `java.util.Stack; ` ` `  `public` `class` `PrintPre { ` ` `  `    ``static` `int` `postIndex; ` ` `  `    ``// Fills preorder traversal of tree with given ` `    ``// inorder and postorder traversals in a stack ` `    ``void` `fillPre(``int``[] in, ``int``[] post, ``int` `inStrt, ` `                 ``int` `inEnd, Stack s) ` `    ``{ ` `        ``if` `(inStrt > inEnd) ` `            ``return``; ` ` `  `        ``// Find index of next item in postorder traversal in ` `        ``// inorder. ` `        ``int` `val = post[postIndex]; ` `        ``int` `inIndex = search(in, val); ` `        ``postIndex--; ` ` `  `        ``// traverse right tree ` `        ``fillPre(in, post, inIndex + ``1``, inEnd, s); ` ` `  `        ``// traverse left tree ` `        ``fillPre(in, post, inStrt, inIndex - ``1``, s); ` ` `  `        ``s.push(val); ` `    ``} ` ` `  `    ``// This function basically initializes postIndex ` `    ``// as last element index, then fills stack with ` `    ``// reverse preorder traversal using printPre ` `    ``void` `printPreMain(``int``[] in, ``int``[] post) ` `    ``{ ` `        ``int` `len = in.length; ` `        ``postIndex = len - ``1``; ` `        ``Stack s = ``new` `Stack(); ` `        ``fillPre(in, post, ``0``, len - ``1``, s); ` `        ``while` `(s.empty() == ``false``) ` `            ``System.out.print(s.pop() + ``" "``); ` `    ``} ` ` `  `    ``// A utility function to search data in in[] ` `    ``int` `search(``int``[] in, ``int` `data) ` `    ``{ ` `        ``int` `i = ``0``; ` `        ``for` `(i = ``0``; i < in.length; i++) ` `            ``if` `(in[i] == data) ` `                ``return` `i; ` `        ``return` `i; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String ars[]) ` `    ``{ ` `        ``int` `in[] = { ``4``, ``10``, ``12``, ``15``, ``18``, ``22``, ``24``, ``25``, ` `                     ``31``, ``35``, ``44``, ``50``, ``66``, ``70``, ``90` `}; ` `        ``int` `post[] = { ``4``, ``12``, ``10``, ``18``, ``24``, ``22``, ``15``, ``31``, ` `                       ``44``, ``35``, ``66``, ``90``, ``70``, ``50``, ``25` `}; ` `        ``PrintPre tree = ``new` `PrintPre(); ` `        ``tree.printPreMain(in, post); ` `    ``} ` `} `

## C#

 `// C# program to print Postorder traversal from given ` `// Inorder and Preorder traversals. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `PrintPre  ` `{ ` ` `  `    ``static` `int` `postIndex; ` ` `  `    ``// Fills preorder traversal of tree with given ` `    ``// inorder and postorder traversals in a stack ` `    ``void` `fillPre(``int``[] a, ``int``[] post, ``int` `inStrt, ` `                ``int` `inEnd, Stack<``int``> s) ` `    ``{ ` `        ``if` `(inStrt > inEnd) ` `            ``return``; ` ` `  `        ``// Find index of next item in postorder traversal in ` `        ``// inorder. ` `        ``int` `val = post[postIndex]; ` `        ``int` `inIndex = search(a, val); ` `        ``postIndex--; ` ` `  `        ``// traverse right tree ` `        ``fillPre(a, post, inIndex + 1, inEnd, s); ` ` `  `        ``// traverse left tree ` `        ``fillPre(a, post, inStrt, inIndex - 1, s); ` ` `  `        ``s.Push(val); ` `    ``} ` ` `  `    ``// This function basically initializes postIndex ` `    ``// as last element index, then fills stack with ` `    ``// reverse preorder traversal using printPre ` `    ``void` `printPreMain(``int``[] a, ``int``[] post) ` `    ``{ ` `        ``int` `len = a.Length; ` `        ``postIndex = len - 1; ` `        ``Stack<``int``> s = ``new` `Stack<``int``>(); ` `        ``fillPre(a, post, 0, len - 1, s); ` `        ``while` `(s.Count!=0) ` `            ``Console.Write(s.Pop() + ``" "``); ` `    ``} ` ` `  `    ``// A utility function to search data in in[] ` `    ``int` `search(``int``[] a, ``int` `data) ` `    ``{ ` `        ``int` `i = 0; ` `        ``for` `(i = 0; i < a.Length; i++) ` `            ``if` `(a[i] == data) ` `                ``return` `i; ` `        ``return` `i; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int` `[]a = { 4, 10, 12, 15, 18, 22, 24, 25, ` `                    ``31, 35, 44, 50, 66, 70, 90 }; ` `        ``int` `[]post = { 4, 12, 10, 18, 24, 22, 15, 31, ` `                    ``44, 35, 66, 90, 70, 50, 25 }; ` `        ``PrintPre tree = ``new` `PrintPre(); ` `        ``tree.printPreMain(a, post); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```25 15 10 4 12 22 18 24 50 35 31 44 70 66 90
```

Time Complexity: The above function visits every node in array. For every visit, it calls search which takes O(n) time. Therefore, overall time complexity of the function is O(n2)

O(n) Solution

We can further optimize above solution to first hash all items of inorder traversal so that we do not have to linearly search items. With hash table available to us, we can search an item in O(1) time.

 `// Java program to print Postorder traversal from given ` `// Inorder and Preorder traversals. ` `import` `java.util.Stack; ` `import` `java.util.HashMap; ` ` `  `public` `class` `PrintPre { ` ` `  `    ``static` `int` `postIndex; ` ` `  `    ``// Fills preorder traversal of tree with given ` `    ``// inorder and postorder traversals in a stack ` `    ``void` `fillPre(``int``[] in, ``int``[] post, ``int` `inStrt, ``int` `inEnd, ` `                 ``Stack s, HashMap hm) ` `    ``{ ` `        ``if` `(inStrt > inEnd) ` `            ``return``; ` ` `  `        ``// Find index of next item in postorder traversal in ` `        ``// inorder. ` `        ``int` `val = post[postIndex]; ` `        ``int` `inIndex = hm.get(val); ` `        ``postIndex--; ` ` `  `        ``// traverse right tree ` `        ``fillPre(in, post, inIndex + ``1``, inEnd, s, hm); ` ` `  `        ``// traverse left tree ` `        ``fillPre(in, post, inStrt, inIndex - ``1``, s, hm); ` ` `  `        ``s.push(val); ` `    ``} ` ` `  `    ``// This function basically initializes postIndex ` `    ``// as last element index, then fills stack with ` `    ``// reverse preorder traversal using printPre ` `    ``void` `printPreMain(``int``[] in, ``int``[] post) ` `    ``{ ` `        ``int` `len = in.length; ` `        ``postIndex = len - ``1``; ` `        ``Stack s = ``new` `Stack(); ` ` `  `        ``// Insert values in a hash map and their indexes. ` `        ``HashMap hm = ` `                     ``new` `HashMap(); ` `        ``for` `(``int` `i = ``0``; i < in.length; i++) ` `            ``hm.put(in[i], i); ` ` `  `        ``// Fill preorder traversal in a stack ` `        ``fillPre(in, post, ``0``, len - ``1``, s, hm); ` ` `  `        ``// Print contents of stack ` `        ``while` `(s.empty() == ``false``) ` `            ``System.out.print(s.pop() + ``" "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String ars[]) ` `    ``{ ` `        ``int` `in[] = { ``4``, ``10``, ``12``, ``15``, ``18``, ``22``, ``24``, ``25``, ` `                     ``31``, ``35``, ``44``, ``50``, ``66``, ``70``, ``90` `}; ` `        ``int` `post[] = { ``4``, ``12``, ``10``, ``18``, ``24``, ``22``, ``15``, ``31``, ` `                       ``44``, ``35``, ``66``, ``90``, ``70``, ``50``, ``25` `}; ` `        ``PrintPre tree = ``new` `PrintPre(); ` `        ``tree.printPreMain(in, post); ` `    ``} ` `} `

Output:

```25 15 10 4 12 22 18 24 50 35 31 44 70 66 90
```

Time Complexity: O(n)

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