We have discussed a simple iterative postorder traversal using two stacks in the previous post. In this post, an approach with only one stack is discussed.
The idea is to move down to leftmost node using left pointer. While moving down, push root and root’s right child to stack. Once we reach leftmost node, print it if it doesn’t have a right child. If it has a right child, then change root so that the right child is processed before.
Following is detailed algorithm.
1.1 Create an empty stack
2.1 Do following while root is not NULL
a) Push root's right child and then root to stack.
b) Set root as root's left child.
2.2 Pop an item from stack and set it as root.
a) If the popped item has a right child and the right child
is at top of stack, then remove the right child from stack,
push the root back and set root as root's right child.
b) Else print root's data and set root as NULL.
2.3 Repeat steps 2.1 and 2.2 while stack is not empty.
Let us consider the following tree
Following are the steps to print postorder traversal of the above tree using one stack.
1. Right child of 1 exists.
Push 3 to stack. Push 1 to stack. Move to left child.
Stack: 3, 1
2. Right child of 2 exists.
Push 5 to stack. Push 2 to stack. Move to left child.
Stack: 3, 1, 5, 2
3. Right child of 4 doesn't exist. '
Push 4 to stack. Move to left child.
Stack: 3, 1, 5, 2, 4
4. Current node is NULL.
Pop 4 from stack. Right child of 4 doesn't exist.
Print 4. Set current node to NULL.
Stack: 3, 1, 5, 2
5. Current node is NULL.
Pop 2 from stack. Since right child of 2 equals stack top element,
pop 5 from stack. Now push 2 to stack.
Move current node to right child of 2 i.e. 5
Stack: 3, 1, 2
6. Right child of 5 doesn't exist. Push 5 to stack. Move to left child.
Stack: 3, 1, 2, 5
7. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist.
Print 5. Set current node to NULL.
Stack: 3, 1, 2
8. Current node is NULL. Pop 2 from stack.
Right child of 2 is not equal to stack top element.
Print 2. Set current node to NULL.
Stack: 3, 1
9. Current node is NULL. Pop 1 from stack.
Since right child of 1 equals stack top element, pop 3 from stack.
Now push 1 to stack. Move current node to right child of 1 i.e. 3
Stack: 1
10. Repeat the same as above steps and Print 6, 7 and 3.
Pop 1 and Print 1.
C++
// C++ program for iterative postorder traversal using one// stack#include <bits/stdc++.h>using namespace std;// A tree nodestruct Node { int data; struct Node *left, *right;};// A utility function to create a new tree nodestruct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return node;}// An iterative function to do postorder traversal of a// given binary treevector<int> postOrderIterative(struct Node* root){ vector<int> postOrderList; // Check for empty tree if (root == NULL) return postOrderList; stack<Node*> S; S.push(root); Node* prev = NULL; while (!S.empty()) { auto current = S.top(); /* go down the tree in search of a leaf an if so process it and pop stack otherwise move down */ if (prev == NULL || prev->left == current || prev->right == current) { if (current->left) S.push(current->left); else if (current->right) S.push(current->right); else { S.pop(); postOrderList.push_back(current->data); } /* go up the tree from left node, if the child is right push it onto stack otherwise process parent and pop stack */ } else if (current->left == prev) { if (current->right) S.push(current->right); else { S.pop(); postOrderList.push_back(current->data); } /* go up the tree from right node and after coming back from right node process parent and pop stack */ } else if (current->right == prev) { S.pop(); postOrderList.push_back(current->data); } prev = current; } return postOrderList;}// Driver program to test above functionsint main(){ // Let us construct the tree shown in above figure struct Node* root = NULL; root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); printf("Post order traversal of binary tree is :\n"); printf("["); vector<int> postOrderList = postOrderIterative(root); for (auto it : postOrderList) cout << it << " "; printf("]"); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program for iterative postorder traversal using one// stack#include <stdio.h>#include <stdlib.h>// Maximum stack size#define MAX_SIZE 100// A tree nodestruct Node { int data; struct Node *left, *right;};// Stack typestruct Stack { int size; int top; struct Node** array;};// A utility function to create a new tree nodestruct Node* newNode(int data){ struct Node* node = (struct Node*)malloc(sizeof(struct Node)); node->data = data; node->left = node->right = NULL; return node;}// A utility function to create a stack of given sizestruct Stack* createStack(int size){ struct Stack* stack = (struct Stack*)malloc(sizeof(struct Stack)); stack->size = size; stack->top = -1; stack->array = (struct Node**)malloc( stack->size * sizeof(struct Node*)); return stack;}// BASIC OPERATIONS OF STACKint isFull(struct Stack* stack){ return stack->top - 1 == stack->size;}int isEmpty(struct Stack* stack){ return stack->top == -1;}void push(struct Stack* stack, struct Node* node){ if (isFull(stack)) return; stack->array[++stack->top] = node;}struct Node* pop(struct Stack* stack){ if (isEmpty(stack)) return NULL; return stack->array[stack->top--];}struct Node* peek(struct Stack* stack){ if (isEmpty(stack)) return NULL; return stack->array[stack->top];}// An iterative function to do postorder traversal of a// given binary treevoid postOrderIterative(struct Node* root){ // Check for empty tree if (root == NULL) return; struct Stack* stack = createStack(MAX_SIZE); do { // Move to leftmost node while (root) { // Push root's right child and then root to // stack. if (root->right) push(stack, root->right); push(stack, root); // Set root as root's left child root = root->left; } // Pop an item from stack and set it as root root = pop(stack); // If the popped item has a right child and the // right child is not processed yet, then make sure // right child is processed before root if (root->right && peek(stack) == root->right) { pop(stack); // remove right child from stack push(stack, root); // push root back to stack root = root->right; // change root so that the // right child is processed // next } else // Else print root's data and set root as NULL { printf("%d ", root->data); root = NULL; } } while (!isEmpty(stack));}// Driver program to test above functionsint main(){ // Let us construct the tree shown in above figure struct Node* root = NULL; root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); printf("Post order traversal of binary tree is :\n"); printf("["); postOrderIterative(root); printf("]"); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// A java program for iterative postorder traversal using// stackimport java.util.ArrayList;import java.util.Stack;// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right; }}class BinaryTree { Node root; ArrayList<Integer> list = new ArrayList<Integer>(); // An iterative function to do postorder traversal // of a given binary tree ArrayList<Integer> postOrderIterative(Node node) { Stack<Node> S = new Stack<Node>(); // Check for empty tree if (node == null) return list; S.push(node); Node prev = null; while (!S.isEmpty()) { Node current = S.peek(); /* go down the tree in search of a leaf an if so process it and pop stack otherwise move down */ if (prev == null || prev.left == current || prev.right == current) { if (current.left != null) S.push(current.left); else if (current.right != null) S.push(current.right); else { S.pop(); list.add(current.data); } /* go up the tree from left node, if the child is right push it onto stack otherwise process parent and pop stack */ } else if (current.left == prev) { if (current.right != null) S.push(current.right); else { S.pop(); list.add(current.data); } /* go up the tree from right node and after coming back from right node process parent and pop stack */ } else if (current.right == prev) { S.pop(); list.add(current.data); } prev = current; } return list; } // Driver program to test above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); ArrayList<Integer> mylist = tree.postOrderIterative(tree.root); System.out.println( "Post order traversal of binary tree is :"); System.out.println(mylist); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 program for iterative postorder traversal # using one stack # Stores the answer ans = [] # A Binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = Nonedef peek(stack): if len(stack) > 0: return stack[-1] return None# A iterative function to do postorder traversal of # a given binary tree def postOrderIterative(root): # Check for empty tree if root is None: return stack = [] while(True): while (root): # Push root's right child and then root to stack if root.right is not None: stack.append(root.right) stack.append(root) # Set root as root's left child root = root.left # Pop an item from stack and set it as root root = stack.pop() # If the popped item has a right child and the # right child is not processed yet, then make sure # right child is processed before root if (root.right is not None and peek(stack) == root.right): stack.pop() # Remove right child from stack stack.append(root) # Push root back to stack root = root.right # change root so that the # right childis processed next # Else print root's data and set root as None else: ans.append(root.data) root = None if (len(stack) <= 0): break# Driver program to test above function root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.left = Node(6) root.right.right = Node(7) print("Post Order traversal of binary tree is")postOrderIterative(root) print(ans)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// A C# program for iterative postorder traversal using stack using System;using System.Collections.Generic;// A binary tree node class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right; } } public class BinaryTree { Node root; List<int> list = new List<int>(); // An iterative function to do postorder traversal // of a given binary tree List<int> postOrderIterative(Node node) { Stack<Node> S = new Stack<Node>(); // Check for empty tree if (node == null) return list; S.Push(node); Node prev = null; while (S.Count != 0) { Node current = S.Peek(); /* go down the tree in search of a leaf an if so process it and pop stack otherwise move down */ if (prev == null || prev.left == current || prev.right == current) { if (current.left != null) S.Push(current.left); else if (current.right != null) S.Push(current.right); else { S.Pop(); list.Add(current.data); } /* go up the tree from left node, if the child is right push it onto stack otherwise process parent and pop stack */ } else if (current.left == prev) { if (current.right != null) S.Push(current.right); else { S.Pop(); list.Add(current.data); } /* go up the tree from right node and after coming back from right node process parent and pop stack */ } else if (current.right == prev) { S.Pop(); list.Add(current.data); } prev = current; } return list; } // Driver code public static void Main(String []args) { BinaryTree tree = new BinaryTree(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); List<int> mylist = tree.postOrderIterative(tree.root); Console.WriteLine("Post order traversal of binary tree is :"); foreach(int i in mylist) Console.Write(i+" "); } } // This code contributed by shikhasingrajput |
Javascript
<script>// A javascript program for iterative postorder traversal using stack// A binary tree node class Node { constructor(item) { this.data=item; this.left=null; this.right=null; } } let root; let list = []; // An iterative function to do postorder traversal // of a given binary tree function postOrderIterative(node) { let S = []; // Check for empty tree if (node == null) return list; S.push(node); let prev = null; while (S.length!=0) { let current = S[S.length-1]; /* go down the tree in search of a leaf an if so process it and pop stack otherwise move down */ if (prev == null || prev.left == current || prev.right == current) { if (current.left != null) S.push(current.left); else if (current.right != null) S.push(current.right); else { S.pop(); list.push(current.data); } /* go up the tree from left node, if the child is right push it onto stack otherwise process parent and pop stack */ } else if (current.left == prev) { if (current.right != null) S.push(current.right); else { S.pop(); list.push(current.data); } /* go up the tree from right node and after coming back from right node process parent and pop stack */ } else if (current.right == prev) { S.pop(); list.push(current.data); } prev = current; } return list; } // Driver program to test above functions // Let us create trees shown in above diagram root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.left = new Node(6); root.right.right = new Node(7); let mylist = postOrderIterative(root); document.write("Post order traversal of binary tree is :<br>"); for(let i = 0; i < mylist.length; i++) { document.write(mylist[i]+" "); } // This code is contributed by unknown2108</script> |
Output
Post order traversal of binary tree is : [4 5 2 6 7 3 1 ]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2:
Push directly root node two times while traversing to the left. While popping if you find stack top() is same as root then go for root->right else print root.
C++
// C++ program for iterative postorder traversal using one// stack#include <bits/stdc++.h>using namespace std;// A tree nodestruct Node { int data; struct Node *left, *right;};// A utility function to create a new tree nodestruct Node* newNode(int data){ struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return node;}// An iterative function to do postorder traversal of a// given binary treevector<int> postOrderIterative(struct Node* root){ vector<int> postOrderList; stack<Node*> st; while (true) { while (root) { st.push(root); st.push(root); root = root->left; } if (st.empty()) return postOrderList; root = st.top(); st.pop(); if (!st.empty() && st.top() == root) root = root->right; else { postOrderList.push_back(root->data); root = NULL; } } return postOrderList;}// Driver program to test above functionsint main(){ // Let us construct the tree shown in above figure struct Node* root = NULL; root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(6); root->right->right = newNode(7); printf("Post order traversal of binary tree is :\n"); printf("["); vector<int> postOrderList = postOrderIterative(root); for (auto it : postOrderList) cout << it << " "; printf("]"); return 0;}// This code is contributed by Sania Kumari Gupta// (kriSania804) |
Java
// Simple Java program to print PostOrder Traversal(Iterative)import java.util.Stack;// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right; }}// create a postorder classclass PostOrder{ Node root; // An iterative function to do postorder traversal // of a given binary tree private void postOrderIterative(Node root) { Stack<Node> stack = new Stack<>(); while(true) { while(root != null) { stack.push(root); stack.push(root); root = root.left; } // Check for empty stack if(stack.empty()) return; root = stack.pop(); if(!stack.empty() && stack.peek() == root) root = root.right; else { System.out.print(root.data + " "); root = null; } } } // Driver program to test above functions public static void main(String args[]) { PostOrder tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); System.out.println("Post order traversal of binary tree is :"); tree.postOrderIterative(tree.root); }} |
Python3
# Simple Python3 program to print # PostOrder Traversal(Iterative)# A binary tree nodeclass Node: def __init__(self, x): self.data = x self.right = None self.left = None# Create a postorder class# An iterative function to do postorder # traversal of a given binary treedef postOrderIterative(root): stack = [] while(True): while(root != None): stack.append(root) stack.append(root) root = root.left # Check for empty stack if (len(stack) == 0): return root = stack.pop() if (len(stack) > 0 and stack[-1] == root): root = root.right else: print(root.data, end = " ") root = None# Driver codeif __name__ == '__main__': # Let us create trees shown # in above diagram root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.left = Node(6) root.right.right = Node(7) print("Post order traversal of binary tree is :") postOrderIterative(root)# This code is contributed by mohit kumar 29 |
C#
// Simple C# program to print PostOrder Traversal(Iterative)using System;using System.Collections.Generic;// A binary tree nodepublic class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right; } }// create a postorder classpublic class PostOrder{ Node root; // An iterative function to do postorder traversal // of a given binary tree private void postOrderIterative(Node root) { Stack<Node> stack = new Stack<Node>(); while(true) { while(root != null) { stack.Push(root); stack.Push(root); root = root.left; } // Check for empty stack if(stack.Count == 0) return; root = stack.Pop(); if(stack.Count != 0 && stack.Peek() == root) root = root.right; else { Console.Write(root.data + " "); root = null; } } } // Driver program to test above functions public static void Main(String []args) { PostOrder tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); Console.WriteLine("Post order traversal of binary tree is :"); tree.postOrderIterative(tree.root); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // Simple JavaScript program to print // PostOrder Traversal(Iterative) // A binary tree node class Node { constructor(item) { this.data = item; this.left = null; this.right = null; } } // create a postorder class class PostOrder { constructor() { this.root = null; } // An iterative function to do postorder traversal // of a given binary tree postOrderIterative(root) { var stack = []; while (true) { while (root != null) { stack.push(root); stack.push(root); root = root.left; } // Check for empty stack if (stack.length == 0) return; root = stack.pop(); if (stack.length != 0 && stack[stack.length - 1] == root) root = root.right; else { document.write(root.data + " "); root = null; } } } } // Driver program to test above functions var tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); document.write("Post order traversal of binary tree is :<br>"); tree.postOrderIterative(tree.root); </script> |
Output
Post order traversal of binary tree is : 4 5 2 6 7 3 1
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3 (Iterative PostOrder Traversal Using Stack and Hashing) :
- Create a Stack for finding the postorder traversal and an unordered map for hashing to mark the visited nodes.
- Initially push the root node in the stack and follow the below steps until the stack is not empty. The stack will get empty when postorder traversal is stored in our answer container data structure.
- Mark the current node (node on the top of stack) as visited in our hashtable.
- If the left child of the current node is not NULL and not visited then push it into the stack.
- Otherwise, if the right child of the top node is not NULL and not visited push it into the stack
- If none of the above two conditions holds true then add the value of the current node to our answer and remove(pop) the current node from the stack.
- When the stack gets empty, we will have postorder traversal stored in our answer data structure (array or vector).
C++
// A C++ program for iterative postorder traversal using// stack#include <bits/stdc++.h>using namespace std;#define MAX_HEIGHT 100000// Tree Nodestruct Node { int data; Node* left; Node* right;};// Utility function to create a new Tree NodeNode* newNode(int val){ Node* temp = new Node; temp->data = val; temp->left = NULL; temp->right = NULL; return temp;}// Function to Build TreeNode* buildTree(string str){ // Corner Case if (str.length() == 0 || str[0] == 'N') return NULL; // Creating vector of strings from input // string after splitting by space vector<string> ip; istringstream iss(str); for (string str; iss >> str;) ip.push_back(str); // Create the root of the tree Node* root = newNode(stoi(ip[0])); // Push the root to the queue queue<Node*> queue; queue.push(root); // Starting from the second element int i = 1; while (!queue.empty() && i < ip.size()) { // Get and remove the front of the queue Node* currNode = queue.front(); queue.pop(); // Get the current node's value from the string string currVal = ip[i]; // If the left child is not null if (currVal != "N") { // Create the left child for the current node currNode->left = newNode(stoi(currVal)); // Push it to the queue queue.push(currNode->left); } // For the right child i++; if (i >= ip.size()) break; currVal = ip[i]; // If the right child is not null if (currVal != "N") { // Create the right child for the current node currNode->right = newNode(stoi(currVal)); // Push it to the queue queue.push(currNode->right); } i++; } return root;}// An iterative function to do postorder traversal// of a given binary treevector<int> postOrder(Node* node){ stack<Node*> s; // vector to store the postorder traversal vector<int> post; // Using unordered map as hash table for hashing to mark // the visited nodes unordered_map<Node*, int> vis; // push the root node in the stack to traverse the tree s.push(node); // stack will be empty when traversal is completed while (!s.empty()) { // mark the node on the top of stack as visited vis[s.top()] = 1; // if left child of the top node is not NULL and not // visited push it into the stack if (s.top()->left != 0) { if (!vis[s.top()->left]) { s.push(s.top()->left); continue; } } // Otherwise if the right child of the top node is // not NULL and not visited push it into the stack if (s.top()->right != 0) { if (!vis[s.top()->right]) { s.push(s.top()->right); continue; } } // Add the value of the top node in our postorder // traversal answer if none of the above two // conditions are met post.push_back(s.top()->data); // Remove the top node from the stack s.pop(); } // post will now contain the postorder traversal of the // tree return post;}int main(){ // Constructing the tree as shown in above diagram string s = "1 2 3 4 5 6 7"; Node* root = buildTree(s); vector<int> ans; ans = postOrder(root); cout << "Post order traversal of binary tree is :\n"; for (int i = 0; i < ans.size(); i++) cout << ans[i] << " "; cout << endl; return 0;}// This code is contributed by Ishan Khandelwal |
Java
// Simple Java program to print PostOrder Traversal(Iterative)import java.util.Stack;// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right; }}// create a postorder classclass PostOrder{ Node root; // An iterative function to do postorder traversal // of a given binary tree private void postOrderIterative(Node root) { Stack<Node> stack = new Stack<>(); while(true) { while(root != null) { stack.push(root); stack.push(root); root = root.left; } // Check for empty stack if(stack.empty()) return; root = stack.pop(); if(!stack.empty() && stack.peek() == root) root = root.right; else { System.out.print(root.data + " "); root = null; } } } // Driver program to test above functions public static void main(String args[]) { PostOrder tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); System.out.println("Post order traversal of binary tree is :"); tree.postOrderIterative(tree.root); }}// This code is contributed by ishankhandelwals. |
Python3
# Simple Python3 program to print# PostOrder Traversal(Iterative)# A binary tree nodeclass Node: def __init__(self, x): self.data = x self.right = None self.left = None# Create a postorder class# An iterative function to do postorder# traversal of a given binary treedef postOrderIterative(root): stack = [] while(True): while(root != None): stack.append(root) stack.append(root) root = root.left # Check for empty stack if (len(stack) == 0): return root = stack.pop() if (len(stack) > 0 and stack[-1] == root): root = root.right else: print(root.data, end=" ") root = None# Driver codeif __name__ == '__main__': # Let us create trees shown # in above diagram root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.left = Node(6) root.right.right = Node(7) print("Post order traversal of binary tree is :") postOrderIterative(root) # This code is contributed by ishankhandelwals. |
C#
// Simple C# program to print PostOrder Traversal(Iterative)using System;using System.Collections.Generic;// A binary tree nodepublic class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right; }}// create a postorder classpublic class PostOrder { Node root; // An iterative function to do postorder traversal // of a given binary tree private void postOrderIterative(Node root) { Stack<Node> stack = new Stack<Node>(); while (true) { while (root != null) { stack.Push(root); stack.Push(root); root = root.left; } // Check for empty stack if (stack.Count == 0) return; root = stack.Pop(); if (stack.Count != 0 && stack.Peek() == root) root = root.right; else { Console.Write(root.data + " "); root = null; } } } // Driver program to test above functions public static void Main(String[] args) { PostOrder tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); Console.WriteLine( "Post order traversal of binary tree is :"); tree.postOrderIterative(tree.root); }}// This code is contributed by ishankhandelwals. |
Javascript
// Simple JavaScript program to print // PostOrder Traversal(Iterative) // A binary tree node class Node { constructor(item) { this.data = item; this.left = null; this.right = null; } } // create a postorder class class PostOrder { constructor() { this.root = null; } // An iterative function to do postorder traversal // of a given binary tree postOrderIterative(root) { var stack = []; while (true) { while (root != null) { stack.push(root); stack.push(root); root = root.left; } // Check for empty stack if (stack.length == 0) return; root = stack.pop(); if (stack.length != 0 && stack[stack.length - 1] == root) root = root.right; else { console.log(root.data + " "); root = null; } } } } // Driver program to test above functions var tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); console.log("Post order traversal of binary tree is :<br>"); tree.postOrderIterative(tree.root); // This code is contributed by ishankhandelwals. |
Output
Post order traversal of binary tree is : 4 5 2 6 7 3 1
Time complexity: O(n) where n is no of nodes in a binary tree
Auxiliary Space: O(n)
This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Method 4:
- In this method, the node is only pushed once.
- Travel to the extreme left using a loop until null.
- Then loop again with the right of the top element of the stack(if it exists). The loop used for traversing to the extreme left is only used in this step in future.
- If the right node is null, then pop until that sub-branch is popped from the stack(to avoid an infinite loop of continuously adding and popping the same thing).
The reason why this program works is that after traversing to extreme left in the beginning, further the program has two paths of execution. One is when the right node is given the control and the other is when the right node hits null. When the right node is given the control, just traverse to the extreme left. If null is hit, pop till that sub branch is eliminated from the stack. So a boolean variable is used so that when right node is given control, it sets to true and program changes to travel extreme left mode and other cases just keep on popping.
C++
// A C++ program for iterative postorder traversal using// stack#include <bits/stdc++.h>using namespace std;#define MAX_HEIGHT 100000// Tree Nodestruct Node { int data; Node* left; Node* right;};// Utility function to create a new Tree NodeNode* newNode(int val){ Node* temp = new Node; temp->data = val; temp->left = NULL; temp->right = NULL; return temp;}// Function to Build TreeNode* buildTree(string str){ // Corner Case if (str.length() == 0 || str[0] == 'N') return NULL; // Creating vector of strings from input // string after splitting by space vector<string> ip; istringstream iss(str); for (string str; iss >> str;) ip.push_back(str); // Create the root of the tree Node* root = newNode(stoi(ip[0])); // Push the root to the queue queue<Node*> queue; queue.push(root); // Starting from the second element int i = 1; while (!queue.empty() && i < ip.size()) { // Get and remove the front of the queue Node* currNode = queue.front(); queue.pop(); // Get the current node's value from the string string currVal = ip[i]; // If the left child is not null if (currVal != "N") { // Create the left child for the current node currNode->left = newNode(stoi(currVal)); // Push it to the queue queue.push(currNode->left); } // For the right child i++; if (i >= ip.size()) break; currVal = ip[i]; // If the right child is not null if (currVal != "N") { // Create the right child for the current node currNode->right = newNode(stoi(currVal)); // Push it to the queue queue.push(currNode->right); } i++; } return root;}// An iterative function to do postorder traversal// of a given binary treevector<int> postOrder(Node* node){ stack<Node*> s; // vector to store the postorder traversal vector<int> post; // Using unordered map as hash table for hashing to mark // the visited nodes unordered_map<Node*, int> vis; // push the root node in the stack to traverse the tree s.push(node); // stack will be empty when traversal is completed while (!s.empty()) { // mark the node on the top of stack as visited vis[s.top()] = 1; // if left child of the top node is not NULL and not // visited push it into the stack if (s.top()->left != 0) { if (!vis[s.top()->left]) { s.push(s.top()->left); continue; } } // Otherwise if the right child of the top node is // not NULL and not visited push it into the stack if (s.top()->right != 0) { if (!vis[s.top()->right]) { s.push(s.top()->right); continue; } } // Add the value of the top node in our postorder // traversal answer if none of the above two // conditions are met post.push_back(s.top()->data); // Remove the top node from the stack s.pop(); } // post will now contain the postorder traversal of the // tree return post;}int main(){ // Constructing the tree as shown in above diagram string s = "1 2 3 4 5 6 7"; Node* root = buildTree(s); vector<int> ans; ans = postOrder(root); cout << "Post order traversal of binary tree is :\n"; for (int i = 0; i < ans.size(); i++) cout << ans[i] << " "; cout << endl; return 0;}// This code is contributed by ishankhandelwals. |
Java
// Simple Java program to print PostOrder Traversal(Iterative)import java.util.Arrays;import java.util.Stack;public class GFG { public static void main (String[] args) { //Constructing the Binary Tree by assigning the right and left //to each nodes //The Node structure of Binary Tree is defined below. Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.left = new Node(6); root.right.right = new Node(7); postorder_iterative(root); } public static void postorder_iterative(Node root){ //The stack to store nodes Stack<Node> s = new Stack<>(); //This check is used to check whether the traversal is from the top element //to it's right. Only then this will be true to invoke the inside while //loop boolean check = true; //Beginning an infinite loop while(true){ //This loop traverses to extreme left node. while(root != null && check){ s.push(root); root = root.left; } //If the stack is empty, the traversal is finished if(s.empty()) break; //To avoid infinite looping this check is necessary //The node of previous iteration is stored in root, so if right //is already visited it won't execute. Hence popping that subtree //to avoid infinite looping. if(root != s.peek().right){ root = s.peek().right; check = true; continue; } //If the node wasn't in any of the special cases above, just pop and //print the value root = s.pop(); System.out.print(root.value + " "); check = false; } }}//Binary Tree Node Structureclass Node{ int value; Node left, right; public Node(int value){ this.value = value; left = right = null; }}//This code is contributed by Ritvik Saran S |
Python
# Python3 program to print the# PostOrder Traversal(Iterative)# Binary tree node structureclass Node: def __init__(self, x): self.data = x self.right = None self.left = None# An iterative function to do postorder # traversal of a given binary treedef postOrderIterative(root): stack = [] check = True while(True): while(root != None): stack.append(root) root = root.left # If the stack is empty, the traversal is finished if (len(stack) == 0): return #To avoid infinite looping this check is necessary if (root != stack[-1].right): root = stack[-1].right check = True continue root = stack.pop() print(root.data, end = " ") check = False# Driver codeif __name__ == '__main__': # Let us create trees shown # in above diagram root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) root.right.left = Node(6) root.right.right = Node(7) print("Post order traversal of binary tree is :") postOrderIterative(root)#This code is contributed by Ritvik Saran S |
C#
// C# program to print PostOrder Traversal(Iterative)using System;using System.Collections.Generic;// Binary tree node structurepublic class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right; }}// Postorder classpublic class PostOrder { Node root; // Iterative function to do postorder traversal private void postOrderIterative(Node root) { Stack<Node> stack = new Stack<Node>(); // This check is used to check whether the traversal // is from the top element to it's right. Only then // this will be true to invoke the inside while loop bool check = true; // Beginning an infinite loop while (true) { // This loop traverses to extreme left node. while (root != null && check) { stack.Push(root); root = root.left; } // If the stack is empty, the traversal is // finished if (stack.Count == 0) return; // To avoid infinite looping this check is // necessary The node of previous iteration is // stored in root, so if right is already visited // it won't execute. Hence popping that subtree // to avoid infinite looping. if (root != stack.Peek().right) { root = stack.Peek().right; check = true; continue; } // If the node wasn't in any of the special // cases above, just pop and print the value root = stack.Pop(); Console.Write(root.data + " "); check = false; } } // Driver program to test above functions public static void Main(String[] args) { PostOrder tree = new PostOrder(); // Let us create trees shown in above diagram tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); Console.WriteLine( "Post order traversal of the binary tree is :"); tree.postOrderIterative(tree.root); }}// This code is contributed by Ritvik Saran S |
Javascript
// Simple JavaScript program to print // PostOrder Traversal(Iterative)// A binary tree nodeclass Node { constructor(item) { this.data = item; this.left = null; this.right = null; }}// create a postorder classclass PostOrder { constructor() { this.root = null; } // An iterative function to do postorder traversal // of a given binary tree postOrderIterative(root) { var stack = []; while (true) { while (root != null) { stack.push(root); stack.push(root); root = root.left; } // Check for empty stack if (stack.length == 0) return; root = stack.pop(); if (stack.length != 0 && stack[stack.length - 1] == root) root = root.right; else { console.log(root.data + " "); root = null; } } }}// Driver program to test above functionsvar tree = new PostOrder();// Let us create trees shown in above diagramtree.root = new Node(1);tree.root.left = new Node(2);tree.root.right = new Node(3);tree.root.left.left = new Node(4);tree.root.left.right = new Node(5);tree.root.right.left = new Node(6);tree.root.right.right = new Node(7);console.log("Post order traversal of binary tree is :<br>");tree.postOrderIterative(tree.root);// This code is contributed by ishankhandelwals. |
Output
4 5 2 6 7 3 1
Time Complexity: O(n), n is the number of nodes of the tree.
Auxiliary Space: O(n), extra stack space is used.
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