Print Postorder traversal from given Inorder and Preorder traversals
AuxiliaryGiven 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 a 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.
Implementation:
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 |
Javascript
<script> // javascript program to print postorder // traversal from preorder and // inorder traversals // A utility function to search x in arr of size n function search(arr , x , n) { for ( var i = 0; i < n; i++) if (arr[i] == x) return i; return -1; } // Prints postorder traversal from // given inorder and preorder traversals function printPostOrder( in1, pre , n) { // The first element in pre is // always root, search it in in // to find left and right subtrees var root = search(in1, pre[0], n); // If left subtree is not empty, // print left subtree if (root != 0) printPostOrder(in1, pre.slice(1, n), root); // If right subtree is not empty, // print right subtree if (root != n - 1) printPostOrder(in1.slice(root+1, n), pre.slice(1+root, n), n - root - 1); // Print root document.write( pre[0] + " " ); } // Driver code var in1 = [ 4, 2, 5, 1, 3, 6 ]; var pre = [ 1, 2, 4, 5, 3, 6 ]; var n = in1.length; document.write( "Postorder traversal <br>" ); printPostOrder(in1, pre, n); // This code contributed by shikhasingrajput </script> |
Postorder traversal 4 5 2 6 3 1
Time Complexity: O(N), where N is the number of nodes in the binary tree. We process every node once, so the time complexity is linear.
Auxiliary Space: O(N). We need to store the recursive calls in the system stack, and in the worst case, the tree is skewed and the height is N. So we need O(N) space for the system stack.
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 ); } } |
Python3
# Python3 program to print Postorder # traversal from given Inorder # and Preorder traversals. preIndex = 0 def search(arr, startIn, endIn, data): i = 0 for i in range (startIn, endIn + 1 ): if (arr[i] = = data): return i return i def printPost(arr, pre, inStrt, inEnd): if (inStrt > inEnd): return # Find index of next item in preorder # traversal in inorder. global preIndex preIndex + = 1 inIndex = search(arr, inStrt, inEnd, pre[preIndex - 1 ]) # 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 print (arr[inIndex], end = " " ) # Driver code arr = [ 4 , 2 , 5 , 1 , 3 , 6 ] pre = [ 1 , 2 , 4 , 5 , 3 , 6 ] len = len (arr) printPost(arr, pre, 0 , len - 1 ) # This code is contributed by Abhijeet Kumar(abhijeet19403) |
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 |
Javascript
<script> // JavaScript program to print Postorder // traversal from given Inorder // and Preorder traversals. class GFG { constructor() { this .preIndex = 0; } printPost(arr, pre, inStrt, inEnd) { if (inStrt > inEnd) { return ; } // Find index of next item in preorder // traversal in inorder. var inIndex = this .search(arr, inStrt, inEnd, pre[ this .preIndex++]); // traverse left tree this .printPost(arr, pre, inStrt, inIndex - 1); // traverse right tree this .printPost(arr, pre, inIndex + 1, inEnd); // print root node at the end of traversal document.write(arr[inIndex] + " " ); } search(arr, startIn, endIn, data) { var i = 0; for (i = startIn; i < endIn; i++) { if (arr[i] == data) { return i; } } return i; } } // Driver code var arr = [4, 2, 5, 1, 3, 6]; var pre = [1, 2, 4, 5, 3, 6]; var len = arr.length; var tree = new GFG(); tree.printPost(arr, pre, 0, len - 1); </script> |
4 5 2 6 3 1
Time Complexity: The above function visits every node in the array. For every visit, it calls search which takes O(n) time. Therefore, overall time complexity of the function is O(n2)
Auxiliary Space: O(1)
Implementation: The above solution can be optimized using hashing. We use a HashMap to store elements and their indexes so that we can quickly find the 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 print Postorder 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) # print root 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 |
Javascript
<script> // Javascript program to print // Postorder traversal from given // Inorder and Preorder traversals. let preIndex = 0; function printPost(In, pre, inStrt, inEnd, hm) { if (inStrt > inEnd) return ; // Find index of next item in // preorder traversal in inorder. let 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 document.write(In[inIndex] + " " ); } function printPostMain(In, pre) { let n = pre.length; let hm = new Map(); for (let i = 0; i < n; i++) hm.set(In[i], i); printPost(In, pre, 0, n - 1, hm); } // Driver code let In = [ 4, 2, 5, 1, 3, 6 ]; let pre = [ 1, 2, 4, 5, 3, 6 ]; printPostMain(In, pre); // This code is contributed by unknown2108 </script> |
4 5 2 6 3 1
Time complexity: O(n)
Auxiliary Space : O(n)
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