Postorder traversal of Binary Tree without recursion and without stack
Prerequisite – Inorder/preorder/postorder traversal of tree
Given a binary tree, perform postorder traversal.
We have discussed below methods for postorder traversal.
1) Recursive Postorder Traversal.
2) Postorder traversal using Stack.
2) Postorder traversal using two Stacks.
In this method a DFS based solution is discussed. We keep track of visited nodes in a hash table.
C++
// CPP program or postorder traversal#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right;};/* Helper function that allocates a new node with thegiven data and NULL left and right pointers. */void postorder(struct Node* head){ struct Node* temp = head; unordered_set<Node*> visited; while (temp && visited.find(temp) == visited.end()) { // Visited left subtree if (temp->left && visited.find(temp->left) == visited.end()) temp = temp->left; // Visited right subtree else if (temp->right && visited.find(temp->right) == visited.end()) temp = temp->right; // Print node else { printf("%d ", temp->data); visited.insert(temp); temp = head; } }}struct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; return (node);}/* Driver program to test above functions*/int main(){ struct Node* root = newNode(8); root->left = newNode(3); root->right = newNode(10); root->left->left = newNode(1); root->left->right = newNode(6); root->left->right->left = newNode(4); root->left->right->right = newNode(7); root->right->right = newNode(14); root->right->right->left = newNode(13); postorder(root); return 0;} |
Java
// JAVA program or postorder traversalimport java.util.*;/* A binary tree node has data, pointer to left childand a pointer to right child */ class Node { int data; Node left, right; Node(int data) { this.data = data; this.left = this.right = null; }};class GFG{ Node root; /* Helper function that allocates a new node with thegiven data and null left and right pointers. */ void postorder(Node head){ Node temp = root; HashSet<Node> visited = new HashSet<>(); while ((temp != null && !visited.contains(temp))) { // Visited left subtree if (temp.left != null && !visited.contains(temp.left)) temp = temp.left; // Visited right subtree else if (temp.right != null && !visited.contains(temp.right)) temp = temp.right; // Print node else { System.out.printf("%d ", temp.data); visited.add(temp); temp = head; } }}/* Driver program to test above functions*/public static void main(String[] args){ GFG gfg = new GFG(); gfg.root = new Node(8); gfg.root.left = new Node(3); gfg.root.right = new Node(10); gfg.root.left.left = new Node(1); gfg.root.left.right = new Node(6); gfg.root.left.right.left = new Node(4); gfg.root.left.right.right = new Node(7); gfg.root.right.right = new Node(14); gfg.root.right.right.left = new Node(13); gfg.postorder(gfg.root);}}// This code is contributed by Rajput-Ji |
Python
# Python program or postorder traversal ''' A binary tree node has data, pointer to left child and a pointer to right child '''class newNode: # Constructor to create a newNode def __init__(self, data): self.data = data self.left = None self.right = None''' Helper function that allocates a new node with the given data and NULL left and right pointers. '''def postorder(head): temp = head visited = set() while (temp and temp not in visited): # Visited left subtree if (temp.left and temp.left not in visited): temp = temp.left # Visited right subtree elif (temp.right and temp.right not in visited): temp = temp.right # Print node else: print(temp.data, end = " ") visited.add(temp) temp = head ''' Driver program to test above functions'''if __name__ == '__main__': root = newNode(8) root.left = newNode(3) root.right = newNode(10) root.left.left = newNode(1) root.left.right = newNode(6) root.left.right.left = newNode(4) root.left.right.right = newNode(7) root.right.right = newNode(14) root.right.right.left = newNode(13) postorder(root) # This code is contributed by # SHUBHAMSINGH10 |
C#
// C# program or postorder traversalusing System;using System.Collections.Generic;/* A binary tree node has data, pointer to left childand a pointer to right child */public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = this.right = null; } };class GFG{ Node root; /* Helper function that allocates a new node with thegiven data and null left and right pointers. */ void postorder(Node head) { Node temp = root; HashSet<Node> visited = new HashSet<Node>(); while ((temp != null && !visited.Contains(temp))) { // Visited left subtree if (temp.left != null && !visited.Contains(temp.left)) temp = temp.left; // Visited right subtree else if (temp.right != null && !visited.Contains(temp.right)) temp = temp.right; // Print node else { Console.Write(temp.data + " "); visited.Add(temp); temp = head; } } } /* Driver code*/ public static void Main(String[] args) { GFG gfg = new GFG(); gfg.root = new Node(8); gfg.root.left = new Node(3); gfg.root.right = new Node(10); gfg.root.left.left = new Node(1); gfg.root.left.right = new Node(6); gfg.root.left.right.left = new Node(4); gfg.root.left.right.right = new Node(7); gfg.root.right.right = new Node(14); gfg.root.right.right.left = new Node(13); gfg.postorder(gfg.root); }}// This code is contributed by Rajput-Ji |
Output:
1 4 7 6 3 13 14 10 8
Alternate Solution:
We can keep visited flag with every node instead of separate hash table.
C++
// CPP program or postorder traversal#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right; bool visited;};void postorder(struct Node* head){ struct Node* temp = head; while (temp && temp->visited == false) { // Visited left subtree if (temp->left && temp->left->visited == false) temp = temp->left; // Visited right subtree else if (temp->right && temp->right->visited == false) temp = temp->right; // Print node else { printf("%d ", temp->data); temp->visited = true; temp = head; } }}struct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; node->visited = false; return (node);}/* Driver program to test above functions*/int main(){ struct Node* root = newNode(8); root->left = newNode(3); root->right = newNode(10); root->left->left = newNode(1); root->left->right = newNode(6); root->left->right->left = newNode(4); root->left->right->right = newNode(7); root->right->right = newNode(14); root->right->right->left = newNode(13); postorder(root); return 0;} |
Java
// Java program or postorder traversalclass GFG{/* A binary tree node has data, pointer to left child and a pointer to right child */static class Node { int data; Node left, right; boolean visited;}static void postorder( Node head){ Node temp = head; while (temp != null && temp.visited == false) { // Visited left subtree if (temp.left != null && temp.left.visited == false) temp = temp.left; // Visited right subtree else if (temp.right != null && temp.right.visited == false) temp = temp.right; // Print node else { System.out.printf("%d ", temp.data); temp.visited = true; temp = head; } }}static Node newNode(int data){ Node node = new Node(); node.data = data; node.left = null; node.right = null; node.visited = false; return (node);}/* Driver code*/public static void main(String []args){ Node root = newNode(8); root.left = newNode(3); root.right = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.left.right.left = newNode(4); root.left.right.right = newNode(7); root.right.right = newNode(14); root.right.right.left = newNode(13); postorder(root);}}// This code is contributed by Arnab Kundu |
Python3
"""Python3 program or postorder traversal """# A Binary Tree Node # Utility function to create a # new tree node class newNode: # Constructor to create a newNode def __init__(self, data): self.data = data self.left = None self.right = None self.visited = Falsedef postorder(head) : temp = head while (temp and temp.visited == False): # Visited left subtree if (temp.left and temp.left.visited == False): temp = temp.left # Visited right subtree elif (temp.right and temp.right.visited == False): temp = temp.right # Print node else: print(temp.data, end = " ") temp.visited = True temp = head # Driver Codeif __name__ == '__main__': root = newNode(8) root.left = newNode(3) root.right = newNode(10) root.left.left = newNode(1) root.left.right = newNode(6) root.left.right.left = newNode(4) root.left.right.right = newNode(7) root.right.right = newNode(14) root.right.right.left = newNode(13) postorder(root)# This code is contributed by # SHUBHAMSINGH10 |
C#
// C# program or postorder traversalusing System;class GFG{/* A binary tree node has data, pointer to left child and a pointer to right child */class Node { public int data; public Node left, right; public bool visited;}static void postorder( Node head){ Node temp = head; while (temp != null && temp.visited == false) { // Visited left subtree if (temp.left != null && temp.left.visited == false) temp = temp.left; // Visited right subtree else if (temp.right != null && temp.right.visited == false) temp = temp.right; // Print node else { Console.Write("{0} ", temp.data); temp.visited = true; temp = head; } }}static Node newNode(int data){ Node node = new Node(); node.data = data; node.left = null; node.right = null; node.visited = false; return (node);}/* Driver code*/public static void Main(String []args){ Node root = newNode(8); root.left = newNode(3); root.right = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.left.right.left = newNode(4); root.left.right.right = newNode(7); root.right.right = newNode(14); root.right.right.left = newNode(13); postorder(root);}}// This code is contributed by 29AjayKumar |
Output:
1 4 7 6 3 13 14 10 8
Time complexity of above solution is O(n2) in worst case we move pointer back to head after visiting every node.
Alternate solution using unordered_map in which we do not have to move pointer back to head, so time complexity is O(n).
CPP
// CPP program or postorder traversal#include <bits/stdc++.h>using namespace std;/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node *left, *right; bool visited;};void postorder(Node* root){ Node* n = root; unordered_map<Node*, Node*> parentMap; parentMap.insert(pair<Node*, Node*>(root, nullptr)); while (n) { if (n->left && parentMap.find(n->left) == parentMap.end()) { parentMap.insert(pair<Node*, Node*>(n->left, n)); n = n->left; } else if (n->right && parentMap.find(n->right) == parentMap.end()) { parentMap.insert(pair<Node*, Node*>(n->right, n)); n = n->right; } else { cout << n->data << " "; n = (parentMap.find(n))->second; } }}struct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; node->visited = false; return (node);}/* Driver program to test above functions*/int main(){ struct Node* root = newNode(8); root->left = newNode(3); root->right = newNode(10); root->left->left = newNode(1); root->left->right = newNode(6); root->left->right->left = newNode(4); root->left->right->right = newNode(7); root->right->right = newNode(14); root->right->right->left = newNode(13); postorder(root); return 0;} |
Output:
1 4 7 6 3 13 14 10 8
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