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Print Postorder traversal from given Inorder and Preorder traversals
  • Difficulty Level : Hard
  • Last Updated : 29 Dec, 2020

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}

Traversals 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++




// 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;
}


Java




// Java program to print postorder
// traversal from preorder and
// inorder traversals
import java.util.Arrays;
 
class GFG
{
 
// A utility function to search x in arr[] of size n
static 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
static void printPostOrder(int in1[],
                    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(in1, pre[0], n);
 
    // If left subtree is not empty,
    // print left subtree
    if (root != 0)
        printPostOrder(in1, Arrays.copyOfRange(pre, 1, n), root);
 
    // If right subtree is not empty,
    // print right subtree
    if (root != n - 1)
        printPostOrder(Arrays.copyOfRange(in1, root+1, n),
            Arrays.copyOfRange(pre, 1+root, n), n - root - 1);
 
    // Print root
    System.out.print( pre[0] + " ");
}
 
// Driver code
public static void main(String args[])
{
    int in1[] = { 4, 2, 5, 1, 3, 6 };
    int pre[] = { 1, 2, 4, 5, 3, 6 };
    int n = in1.length;
    System.out.println("Postorder traversal " );
    printPostOrder(in1, pre, n);
}
}
// This code is contributed by Arnab Kundu


Python3




# Python3 program to print postorder
# traversal from preorder and inorder
# traversals
 
# A utility function to search x in
# arr[] of size n
def search(arr, x, n):
      
    for i in range(n):
        if (arr[i] == x):
            return i
 
    return -1
 
# Prints postorder traversal from
# given inorder and preorder traversals
def printPostOrder(In, pre, n):
     
    # The first element in pre[] is always
    # root, search it in in[] to find left
    # and right subtrees
    root = search(In, pre[0], n)
 
    # If left subtree is not empty,
    # print left subtree
    if (root != 0):
        printPostOrder(In, pre[1:n], root)
 
    # If right subtree is not empty,
    # print right subtree
    if (root != n - 1):
        printPostOrder(In[root + 1 : n],
                      pre[root + 1 : n],
                      n - root - 1)
 
    # Print root
    print(pre[0], end = " ")
 
# Driver code
In = [ 4, 2, 5, 1, 3, 6 ]
pre = [ 1, 2, 4, 5, 3, 6 ]
n = len(In)
 
print("Postorder traversal ")
 
printPostOrder(In, pre, n)
 
# This code is contributed by avanitrachhadiya2155


C#




// C# program to print postorder
// traversal from preorder and
// inorder traversals
using System;
 
class GFG
{
 
// A utility function to search x
// in []arr of size n
static 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
static void printPostOrder(int []in1,
                    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(in1, pre[0], n);
 
    // If left subtree is not empty,
    // print left subtree
    int []ar;
    if (root != 0)
    {
        ar = new int[n - 1];
        Array.Copy(pre, 1, ar, 0, n - 1);
        printPostOrder(in1, ar, root);
    }
     
    // If right subtree is not empty,
    // print right subtree
    if (root != n - 1)
    {
        ar = new int[n - (root + 1)];
        Array.Copy(in1, root + 1, ar, 0,
                        n - (root + 1));
        int []ar1 = new int[n - (root + 1)];
        Array.Copy(pre, root + 1, ar1, 0,
                         n - (root + 1));
        printPostOrder(ar, ar1, n - root - 1);
    }
     
    // Print root
    Console.Write(pre[0] + " ");
}
 
// Driver code
public static void Main(String []args)
{
    int []in1 = { 4, 2, 5, 1, 3, 6 };
    int []pre = { 1, 2, 4, 5, 3, 6 };
    int n = in1.Length;
    Console.WriteLine("Postorder traversal " );
    printPostOrder(in1, pre, n);
}
}
 
// This code is contributed by 29AjayKumar


Output: 

Postorder traversal 
4 5 2 6 3 1

 

Below is the implementation. 

C++




// C++ program to print Postorder
// traversal from given Inorder
// and Preorder traversals.
#include <iostream>
using namespace std;
 
int preIndex = 0;
 
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;
}
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
    cout << arr[inIndex] << " ";
}
 
// Driver code
int main()
{
    int arr[] = {4, 2, 5, 1, 3, 6};
    int pre[] = {1, 2, 4, 5, 3, 6};
    int len = sizeof(arr)/sizeof(arr[0]);
    printPost(arr, pre, 0, len - 1);
}
 
// This code is contributed by SHUBHAMSINGH10


Java




// 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);
    }
}


C#




// 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


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. 

C++




// C++ program to print Postorder traversal from
// given Inorder and Preorder traversals.
#include<bits/stdc++.h>
using namespace std;
 
int preIndex = 0;
void printPost(int in[], int pre[], int inStrt,
               int inEnd, map<int, int> hm)
{
    if (inStrt > inEnd)
        return;        
 
    // Find index of next item in preorder traversal in
    // inorder.
    int inIndex = hm[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
    cout << in[inIndex] << " ";
}
 
void printPostMain(int in[], int pre[],int n)
{
    map<int,int> hm ;
    for (int i = 0; i < n; i++)
    hm[in[i]] = i;
         
    printPost(in, pre, 0, n - 1, hm);
}
 
// Driver code
int main()
{
    int in[] = { 4, 2, 5, 1, 3, 6 };
    int pre[] = { 1, 2, 4, 5, 3, 6 };
    int n = sizeof(pre)/sizeof(pre[0]);
     
    printPostMain(in, pre, n);
    return 0;
}
 
// This code is contributed by Arnab Kundu


Java




// 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);
    }
}


Python3




# Python3 program to prPostorder traversal from
# given Inorder and Preorder traversals.
 
def printPost(inn, pre, inStrt, inEnd):
    global preIndex, hm
    if (inStrt > inEnd):
        return
 
    # Find index of next item in preorder traversal in
    # inorder.
    inIndex = hm[pre[preIndex]]
    preIndex += 1
 
    # traverse left tree
    printPost(inn, pre, inStrt, inIndex - 1)
 
    # traverse right tree
    printPost(inn, pre, inIndex + 1, inEnd)
 
    # prroot node at the end of traversal
    print(inn[inIndex], end = " ")
 
def printPostMain(inn, pre, n):
 
    for i in range(n):
        hm[inn[i]] = i
 
    printPost(inn, pre, 0, n - 1)
 
# Driver code
if __name__ == '__main__':
    hm = {}
    preIndex = 0
    inn = [4, 2, 5, 1, 3, 6]
    pre = [1, 2, 4, 5, 3, 6]
 
    n = len(pre)
 
    printPostMain(inn, pre, n)
 
# This code is contributed by mohit kumar 29


C#




// 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


Output: 

4 5 2 6 3 1

 

Time complexity: O(n)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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