Given an array representing preorder traversal of BST, print its postorder traversal.
Examples:
Input : 40 30 35 80 100 Output : 35 30 100 80 40 Input : 40 30 32 35 80 90 100 120 Output : 35 32 30 120 100 90 80 40
Prerequisite: Construct BST from given preorder traversal
Simple Approach: A simple solution is to first construct BST from a given preorder traversal as described in this post. After constructing the tree, perform postorder traversal on it.
Efficient Approach:
An efficient approach is to find postorder traversal without constructing the tree. The idea is to traverse the given preorder array and maintain a range in which current element should lie. This is to ensure that the BST property is always satisfied. Initially the range is set to {minval = INT_MIN, maxval = INT_MAX}. In preorder traversal, the first element is always the root, and it will certainly lie in the initial range.
So store the first element of the preorder array. In postorder traversal, first left and right subtrees are printed and then root data is printed. So first recursive call for left and right subtrees are performed and then the value of root is printed. For left subtree range is updated to {minval, root->data} and for right subtree range is updated to {root->data, maxval}. If the current preorder array element does not lie in the range specified for it, then it does not belong to a current subtree, return from recursive calls until the correct position of that element is not found.
Below is the implementation of the above approach:
C++
// C++ program for finding postorder// traversal of BST from preorder traversal#include <bits/stdc++.h>using namespace std;// Function to find postorder traversal from// preorder traversal.void findPostOrderUtil(int pre[], int n, int minval, int maxval, int& preIndex){ // If entire preorder array is traversed then // return as no more element is left to be // added to post order array. if (preIndex == n) return; // If array element does not lie in range specified, // then it is not part of current subtree. if (pre[preIndex] < minval || pre[preIndex] > maxval) { return; } // Store current value, to be printed later, after // printing left and right subtrees. Increment // preIndex to find left and right subtrees, // and pass this updated value to recursive calls. int val = pre[preIndex]; preIndex++; // All elements with value between minval and val // lie in left subtree. findPostOrderUtil(pre, n, minval, val, preIndex); // All elements with value between val and maxval // lie in right subtree. findPostOrderUtil(pre, n, val, maxval, preIndex); cout << val << " ";}// Function to find postorder traversal.void findPostOrder(int pre[], int n){ // To store index of element to be // traversed next in preorder array. // This is passed by reference to // utility function. int preIndex = 0; findPostOrderUtil(pre, n, INT_MIN, INT_MAX, preIndex);}// Driver codeint main(){ int pre[] = { 40, 30, 35, 80, 100 }; int n = sizeof(pre) / sizeof(pre[0]); // Calling function findPostOrder(pre, n); return 0;} |
Java
// Java program for finding postorder// traversal of BST from preorder traversalimport java.util.*;class Solution { static class INT { int data; INT(int d) { data = d; } } // Function to find postorder traversal from // preorder traversal. static void findPostOrderUtil(int pre[], int n, int minval, int maxval, INT preIndex) { // If entire preorder array is traversed then // return as no more element is left to be // added to post order array. if (preIndex.data == n) return; // If array element does not lie in range specified, // then it is not part of current subtree. if (pre[preIndex.data] < minval || pre[preIndex.data] > maxval) { return; } // Store current value, to be printed later, after // printing left and right subtrees. Increment // preIndex to find left and right subtrees, // and pass this updated value to recursive calls. int val = pre[preIndex.data]; preIndex.data++; // All elements with value between minval and val // lie in left subtree. findPostOrderUtil(pre, n, minval, val, preIndex); // All elements with value between val and maxval // lie in right subtree. findPostOrderUtil(pre, n, val, maxval, preIndex); System.out.print(val + " "); } // Function to find postorder traversal. static void findPostOrder(int pre[], int n) { // To store index of element to be // traversed next in preorder array. // This is passed by reference to // utility function. INT preIndex = new INT(0); findPostOrderUtil(pre, n, Integer.MIN_VALUE, Integer.MAX_VALUE, preIndex); } // Driver code public static void main(String args[]) { int pre[] = { 40, 30, 35, 80, 100 }; int n = pre.length; // Calling function findPostOrder(pre, n); }}// This code is contributed// by Arnab Kundu |
Python3
"""Python3 program for finding postorder traversal of BST from preorder traversal"""INT_MIN = -2**31INT_MAX = 2**31# Function to find postorder traversal# from preorder traversal.def findPostOrderUtil(pre, n, minval, maxval, preIndex): # If entire preorder array is traversed # then return as no more element is left # to be added to post order array. if (preIndex[0] == n): return # If array element does not lie in # range specified, then it is not # part of current subtree. if (pre[preIndex[0]] < minval or pre[preIndex[0]] > maxval): return # Store current value, to be printed later, # after printing left and right subtrees. # Increment preIndex to find left and right # subtrees, and pass this updated value to # recursive calls. val = pre[preIndex[0]] preIndex[0] += 1 # All elements with value between minval # and val lie in left subtree. findPostOrderUtil(pre, n, minval, val, preIndex) # All elements with value between val # and maxval lie in right subtree. findPostOrderUtil(pre, n, val, maxval, preIndex) print(val, end=" ")# Function to find postorder traversal.def findPostOrder(pre, n): # To store index of element to be # traversed next in preorder array. # This is passed by reference to # utility function. preIndex = [0] findPostOrderUtil(pre, n, INT_MIN, INT_MAX, preIndex)# Driver Codeif __name__ == '__main__': pre = [40, 30, 35, 80, 100] n = len(pre) # Calling function findPostOrder(pre, n)# This code is contributed by# SHUBHAMSINGH10 |
C#
// C# program for finding postorder// traversal of BST from preorder traversalusing System;class GFG { public class INT { public int data; public INT(int d) { data = d; } } // Function to find postorder traversal from // preorder traversal. public static void findPostOrderUtil(int[] pre, int n, int minval, int maxval, INT preIndex) { // If entire preorder array is traversed // then return as no more element is left // to be added to post order array. if (preIndex.data == n) { return; } // If array element does not lie in // range specified, then it is not // part of current subtree. if (pre[preIndex.data] < minval || pre[preIndex.data] > maxval) { return; } // Store current value, to be printed // later, after printing left and right // subtrees. Increment preIndex to find // left and right subtrees, and pass this // updated value to recursive calls. int val = pre[preIndex.data]; preIndex.data++; // All elements with value between // minval and val lie in left subtree. findPostOrderUtil(pre, n, minval, val, preIndex); // All elements with value between // val and maxval lie in right subtree. findPostOrderUtil(pre, n, val, maxval, preIndex); Console.Write(val + " "); } // Function to find postorder traversal. public static void findPostOrder(int[] pre, int n) { // To store index of element to be // traversed next in preorder array. // This is passed by reference to // utility function. INT preIndex = new INT(0); findPostOrderUtil(pre, n, int.MinValue, int.MaxValue, preIndex); } // Driver code public static void Main(string[] args) { int[] pre = new int[] { 40, 30, 35, 80, 100 }; int n = pre.Length; // Calling function findPostOrder(pre, n); }}// This code is contributed by Shrikant13 |
Javascript
<script>// Javascript program for finding postorder// traversal of BST from preorder traversal class INT{ constructor(d) { this.data=d; }}// Function to find postorder traversal from // preorder traversal.function findPostOrderUtil(pre,n,minval,maxval,preIndex){ // If entire preorder array is traversed then // return as no more element is left to be // added to post order array. if (preIndex.data == n) return; // If array element does not lie in range specified, // then it is not part of current subtree. if (pre[preIndex.data] < minval || pre[preIndex.data] > maxval) { return; } // Store current value, to be printed later, after // printing left and right subtrees. Increment // preIndex to find left and right subtrees, // and pass this updated value to recursive calls. let val = pre[preIndex.data]; preIndex.data++; // All elements with value between minval and val // lie in left subtree. findPostOrderUtil(pre, n, minval, val, preIndex); // All elements with value between val and maxval // lie in right subtree. findPostOrderUtil(pre, n, val, maxval, preIndex); document.write(val + " ");} // Function to find postorder traversal.function findPostOrder(pre,n){ // To store index of element to be // traversed next in preorder array. // This is passed by reference to // utility function. let preIndex = new INT(0); findPostOrderUtil(pre, n, Number.MIN_VALUE, Number.MAX_VALUE, preIndex);}// Driver codelet pre=[40, 30, 35, 80, 100];let n = pre.length;// Calling functionfindPostOrder(pre, n); // This code is contributed by unknown2108</script> |
Output
35 30 100 80 40
Complexity Analysis:
- Time Complexity: O(N), where N is the number of nodes.
- Auxiliary Space: O(N) (Function call stack size).
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