Print Postorder traversal from given Inorder and Preorder traversals

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

Example:

Input:
Inorder traversal in[] = {4, 2, 5, 1, 3, 6}
Preorder traversal pre[] = {1, 2, 4, 5, 3, 6}

Output:
Postorder traversal is {4, 5, 2, 6, 3, 1}

Trversals in the above example represents following tree

         1
      /    \    
     2       3
   /   \      \
  4     5      6

A naive method is to first construct the tree, then use simple recursive method to print postorder traversal of the constructed tree.

We can print postorder 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 recursively print left subtree, then recursively print right subtree. Finally, print root. To find boundaries of left and right subtrees in pre[] 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 pre[], 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.



C++

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// C++ program to print postorder traversal from preorder and inorder traversals
#include <iostream>
using namespace std;
  
// A utility function to search x in arr[] of size n
int search(int arr[], int x, int n)
{
    for (int i = 0; i < n; i++)
        if (arr[i] == x)
            return i;
    return -1;
}
  
// Prints postorder traversal from given inorder and preorder traversals
void printPostOrder(int in[], int pre[], int n)
{
    // The first element in pre[] is always root, search it
    // in in[] to find left and right subtrees
    int root = search(in, pre[0], n);
  
    // If left subtree is not empty, print left subtree
    if (root != 0)
        printPostOrder(in, pre + 1, root);
  
    // If right subtree is not empty, print right subtree
    if (root != n - 1)
        printPostOrder(in + root + 1, pre + root + 1, n - root - 1);
  
    // Print root
    cout << pre[0] << " ";
}
  
// Driver program to test above functions
int main()
{
    int in[] = { 4, 2, 5, 1, 3, 6 };
    int pre[] = { 1, 2, 4, 5, 3, 6 };
    int n = sizeof(in) / sizeof(in[0]);
    cout << "Postorder traversal " << endl;
    printPostOrder(in, pre, n);
    return 0;
}

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Python3

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# Python program to print postorder 
# traversal from preorder and 
# inorder traversals
def printpostorder(inorder, preorder, n):
    if preorder[0] in inorder:
        root = inorder.index(preorder[0])
          
    if root != 0: # left subtree exists
        printpostorder(inorder[:root], 
                       preorder[1:root + 1], 
                       len(inorder[:root]))
      
    if root != n - 1: # right subtree exists
        printpostorder(inorder[root + 1:],  
                       preorder[root + 1:], 
                       len(inorder[root + 1:]))
      
    print preorder[0],
          
# Driver Code
inorder = [4, 2, 5, 1, 3, 6];
preorder = [1, 2, 4, 5, 3, 6];
n = len(inorder)
print "Postorder traversal "
printpostorder(inorder, preorder, n)
  
# This code is contributed by SaiNath

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Output:

Postorder traversal 
4 5 2 6 3 1

Below is Java implementation.

Java

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// Java program to print Postorder traversal from given Inorder
// and Preorder traversals.
  
public class PrintPost {
    static int preIndex = 0;
    void printPost(int[] in, int[] pre, int inStrt, int inEnd)
    {
        if (inStrt > inEnd) 
            return;        
  
        // Find index of next item in preorder traversal in
        // inorder.
        int inIndex = search(in, inStrt, inEnd, pre[preIndex++]);
  
        // traverse left tree
        printPost(in, pre, inStrt, inIndex - 1);
  
        // traverse right tree
        printPost(in, pre, inIndex + 1, inEnd);
  
        // print root node at the end of traversal
        System.out.print(in[inIndex] + " ");
    }
  
    int search(int[] in, int startIn, int endIn, int data)
    {
        int i = 0;
        for (i = startIn; i < endIn; i++) 
            if (in[i] == data) 
                return i;            
        return i;
    }
  
    // Driver code
    public static void main(String ars[])
    {
        int in[] = { 4, 2, 5, 1, 3, 6 };
        int pre[] = { 1, 2, 4, 5, 3, 6 };
        int len = in.length;
        PrintPost tree = new PrintPost();
        tree.printPost(in, pre, 0, len - 1);
    }
}

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C#

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// C# program to print Postorder 
// traversal from given Inorder 
// and Preorder traversals. 
using System;
  
class GFG
{
public static int preIndex = 0;
public virtual void printPost(int[] arr, int[] pre, 
                              int inStrt, int inEnd)
{
    if (inStrt > inEnd)
    {
        return;
    }
  
    // Find index of next item in preorder
    // traversal in inorder. 
    int inIndex = search(arr, inStrt, inEnd, 
                         pre[preIndex++]);
  
    // traverse left tree 
    printPost(arr, pre, inStrt, inIndex - 1);
  
    // traverse right tree 
    printPost(arr, pre, inIndex + 1, inEnd);
  
    // print root node at the end of traversal 
    Console.Write(arr[inIndex] + " ");
}
  
public virtual int search(int[] arr, int startIn,
                          int endIn, int data)
{
    int i = 0;
    for (i = startIn; i < endIn; i++)
    {
        if (arr[i] == data)
        {
            return i;
        }
    }
    return i;
}
  
// Driver code 
public static void Main(string[] ars)
{
    int[] arr = new int[] {4, 2, 5, 1, 3, 6};
    int[] pre = new int[] {1, 2, 4, 5, 3, 6};
    int len = arr.Length;
    GFG tree = new GFG();
    tree.printPost(arr, pre, 0, len - 1);
}
}
  
// This code is contributed by Shrikant13

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Output:

4 5 2 6 3 1

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)

The above solution can be optimized using hashing. We use a HashMap to store elements and their indexes so that we can quickly find index of an element.

Java

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// Java program to print Postorder traversal from 
// given Inorder and Preorder traversals. 
import java.util.*;
  
public class PrintPost { 
    static int preIndex = 0
    void printPost(int[] in, int[] pre, int inStrt,
               int inEnd, HashMap<Integer, Integer> hm) 
    
        if (inStrt > inEnd) 
            return;         
  
        // Find index of next item in preorder traversal in 
        // inorder. 
        int inIndex = hm.get(pre[preIndex++]); 
  
        // traverse left tree 
        printPost(in, pre, inStrt, inIndex - 1, hm); 
  
        // traverse right tree 
        printPost(in, pre, inIndex + 1, inEnd, hm); 
  
        // print root node at the end of traversal 
        System.out.print(in[inIndex] + " "); 
    
  
    void printPostMain(int[] in, int[] pre) 
    {
        int n = pre.length;
        HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
        for (int i=0; i<n; i++)
           hm.put(in[i], i);
             
        printPost(in, pre, 0, n-1, hm);
    }
  
    // Driver code 
    public static void main(String ars[]) 
    
        int in[] = { 4, 2, 5, 1, 3, 6 }; 
        int pre[] = { 1, 2, 4, 5, 3, 6 }; 
        PrintPost tree = new PrintPost(); 
        tree.printPostMain(in, pre); 
    

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C#

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// C# program to print Postorder 
// traversal from given Inorder 
// and Preorder traversals. 
using System;
  
class GFG
{
public static int preIndex = 0;
public virtual void printPost(int[] arr, int[] pre, 
                              int inStrt, int inEnd)
{
    if (inStrt > inEnd)
    {
        return;
    }
  
    // Find index of next item in preorder
    // traversal in inorder. 
    int inIndex = search(arr, inStrt, inEnd, 
                         pre[preIndex++]);
  
    // traverse left tree 
    printPost(arr, pre, inStrt, inIndex - 1);
  
    // traverse right tree 
    printPost(arr, pre, inIndex + 1, inEnd);
  
    // print root node at the
    // end of traversal 
    Console.Write(arr[inIndex] + " ");
}
  
public virtual int search(int[] arr, int startIn,
                          int endIn, int data)
{
    int i = 0;
    for (i = startIn; i < endIn; i++)
    {
        if (arr[i] == data)
        {
            return i;
        }
    }
    return i;
}
  
// Driver code 
public static void Main(string[] ars)
{
    int[] arr = new int[] {4, 2, 5, 1, 3, 6};
    int[] pre = new int[] {1, 2, 4, 5, 3, 6};
    int len = arr.Length;
    GFG tree = new GFG();
    tree.printPost(arr, pre, 0, len - 1);
}
}
  
// This code is contributed by Shrikant13

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Output:

4 5 2 6 3 1

Time complexity: O(n)

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Improved By : sainathcvs, shrikanth13



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