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Post Order Traversal of Binary Tree in O(N) using O(1) space

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  • Difficulty Level : Hard
  • Last Updated : 25 Oct, 2021

Prerequisites:- Morris Inorder Traversal, Tree Traversals (Inorder, Preorder and Postorder)
Given a Binary Tree, the task is to print the elements in post order using O(N) time complexity and constant space.

Input:   1 
       /   \
     2       3
    / \     / \
   4   5   6   7
  / \
 8   9
Output: 8 9 4 5 2 6 7 3 1

Input:   5 
       /   \
     7       3
    / \     / \
   4   11  13  9
  / \
 8   4
Output: 8 4 4 11 7 13 9 3 5

Method 1: Using Morris Inorder Traversal

  1. Create a dummy node and make the root as it’s left child.
  2. Initialize current with dummy node.
  3. While current is not NULL 
    • If the current does not have a left child traverse the right child, current = current->right
    • Otherwise, 
      1. Find the rightmost child in the left subtree.
      2. If rightmost child’s right child is NULL
        • Make current as the right child of the rightmost node.
        • Traverse the left child, current = current->left
      3. Otherwise, 
        • Set the rightmost child’s right pointer to NULL.
        • From current’s left child, traverse along with the right children until the rightmost child and reverse the pointers.
        • Traverse back from rightmost child to current’s left child node by reversing the pointers and printing the elements.
        • Traverse the right child, current = current->right

Below is the diagram showing the rightmost child in the left subtree, pointing to its inorder successor. 

Below is the diagram which highlights the path 1->2->5->9 and the way the nodes are processed and printed as per the above algorithm.  

Below is the implementation of the above approach:

C++




// C++ program to implement
// Post Order traversal
// of Binary Tree in O(N)
// time and O(1) space
#include <bits/stdc++.h>
using namespace std;
 
class node
{
    public:
    int data;
    node *left, *right;
};
 
// Helper function that allocates a
// new node with the given data and
// NULL left and right pointers.
node* newNode(int data)
{
    node* temp = new node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Postorder traversal without recursion
// and without stack
void postOrderConstSpace(node* root)
{
    if (root == NULL)
            return;
 
        node* current = newNode(-1);
        node* pre = NULL;
        node* prev = NULL;
        node* succ = NULL;
        node* temp = NULL;
         
        current->left = root;
         
    while (current)
    {
         
        // If left child is null.
        // Move to right child.
        if (current->left == NULL)
        {
            current = current->right;
        }
        else
        {
            pre = current->left;
             
            // Inorder predecessor
            while (pre->right &&
                pre->right != current)
                pre = pre->right;
             
            // The connection between current and
            // predecessor is made
            if (pre->right == NULL)
            {
                 
                // Make current as the right
                // child of the right most node
                pre->right = current;
                 
                // Traverse the left child
                current = current->left;
            }
            else
            {
                pre->right = NULL;
                succ = current;
                current = current->left;
                prev = NULL;
                 
                // Traverse along the right
                // subtree to the
                // right-most child
                while (current != NULL)
                {
                    temp = current->right;
                    current->right = prev;
                    prev = current;
                    current = temp;
                }
                 
                // Traverse back
                // to current's left child
                // node
                while (prev != NULL)
                {
                    cout << prev->data << " ";
                    temp = prev->right;
                    prev->right = current;
                    current = prev;
                    prev = temp;
                }
 
                current = succ;
                current = current->right;
            }
        }
    }
}
 
// Driver code
int main()
{
   /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
      */
    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);
 
    root->left->right->left = newNode(8);
    root->left->right->right = newNode(9);
     
    postOrderConstSpace(root);
    return 0;
}
 
// This code is contributed by Saurav Chaudhary


Java




// Java program to implement
// Post Order traversal
// of Binary Tree in O(N)
// time and O(1) space
 
// Definition of the
// binary tree
class TreeNode {
    public int data;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int data)
    {
        this.data = data;
    }
 
    public String toString()
    {
        return data + " ";
    }
}
 
public class PostOrder {
 
    TreeNode root;
 
    // Function to find Post Order
    // Traversal Using Constant space
    void postOrderConstantspace(TreeNode
                                    root)
    {
        if (root == null)
            return;
 
        TreeNode current
            = new TreeNode(-1),
            pre = null;
        TreeNode prev = null,
                succ = null,
                temp = null;
        current.left = root;
 
        while (current != null) {
 
            // Go to the right child
            // if current does not
            // have a left child
 
            if (current.left == null) {
                current = current.right;
            }
 
            else {
 
                // Traverse left child
                pre = current.left;
 
                // Find the right most child
                // in the left subtree
                while (pre.right != null
                    && pre.right != current)
                    pre = pre.right;
 
                if (pre.right == null) {
 
                    // Make current as the right
                    // child of the right most node
                    pre.right = current;
 
                    // Traverse the left child
                    current = current.left;
                }
 
                else {
                    pre.right = null;
                    succ = current;
                    current = current.left;
                    prev = null;
 
                    // Traverse along the right
                    // subtree to the
                    // right-most child
 
                    while (current != null) {
                        temp = current.right;
                        current.right = prev;
                        prev = current;
                        current = temp;
                    }
 
                    // Traverse back from
                    // right most child to
                    // current's left child node
 
                    while (prev != null) {
 
                        System.out.print(prev);
                        temp = prev.right;
                        prev.right = current;
                        current = prev;
                        prev = temp;
                    }
 
                    current = succ;
                    current = current.right;
                }
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        /* Constructed tree is as follows:-
                      1
                    /   \
                   2     3
                  / \    / \
                 4   5   6  7
                    / \
                    8  9
        */
        PostOrder tree = new PostOrder();
        tree.root = new TreeNode(1);
        tree.root.left = new TreeNode(2);
        tree.root.right = new TreeNode(3);
        tree.root.left.left = new TreeNode(4);
        tree.root.left.right
            = new TreeNode(5);
        tree.root.right.left
            = new TreeNode(6);
        tree.root.right.right
            = new TreeNode(7);
        tree.root.left.right.left
            = new TreeNode(8);
        tree.root.left.right.right
            = new TreeNode(9);
 
        tree.postOrderConstantspace(
            tree.root);
    }
}


Python3




# Python3 program to implement
# Post Order traversal
# of Binary Tree in O(N)
# time and O(1) space
class node:
     
    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
# None left and right pointers.
def newNode(data):
 
    temp = node(data)
    return temp
 
# Postorder traversal without recursion
# and without stack
def postOrderConstSpace(root):
 
    if (root == None):
        return
 
    current = newNode(-1)
    pre = None
    prev = None
    succ = None
    temp = None
     
    current.left = root
         
    while (current):
         
        # If left child is None.
        # Move to right child.
        if (current.left == None):
            current = current.right
        else:
            pre = current.left
             
            # Inorder predecessor
            while (pre.right and
                   pre.right != current):
                pre = pre.right
             
            # The connection between current
            # and predecessor is made
            if (pre.right == None):
                 
                # Make current as the right
                # child of the right most node
                pre.right = current
                 
                # Traverse the left child
                current = current.left
             
            else:
             
                pre.right = None
                succ = current
                current = current.left
                prev = None
                 
                # Traverse along the right
                # subtree to the
                # right-most child
                while (current != None):
                    temp = current.right
                    current.right = prev
                    prev = current
                    current = temp
             
                # Traverse back
                # to current's left child
                # node
                while (prev != None):
                    print(prev.data, end = ' ')
                    temp = prev.right
                    prev.right = current
                    current = prev
                    prev = temp
 
                current = succ
                current = current.right
 
# Driver code
if __name__=='__main__':
     
    ''' Constructed tree is as follows:-
                       1
                    /     \
                   2      3
                  / \     / \
                 4   5  6   7
                    / \
                   8   9
        '''
    root = None
 
    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)
 
    root.left.right.left = newNode(8)
    root.left.right.right = newNode(9)
     
    postOrderConstSpace(root)
 
# This code is contributed by pratham76


C#




// C# program to implement
// Post Order traversal
// of Binary Tree in O(N)
// time and O(1) space
using System;
 
// Definition of the
// binary tree
public class  TreeNode
{
    public int data;
    public TreeNode left, right;
   
    public TreeNode(int item)
    {
        data = item;
        left = right = null;
    }
}
 
class PostOrder{
     
public TreeNode root;
 
// Function to find Post Order
// Traversal Using Constant space
void postOrderConstantspace(TreeNode root)
{
    if (root == null)
        return;
   
    TreeNode current = new TreeNode(-1), pre = null;
    TreeNode prev = null,
             succ = null,
             temp = null;
              
    current.left = root;
 
    while (current != null)
    {
         
        // Go to the right child
        // if current does not
        // have a left child
        if (current.left == null)
        {
            current = current.right;
        }
 
        else
        {
             
            // Traverse left child
            pre = current.left;
 
            // Find the right most child
            // in the left subtree
            while (pre.right != null &&
                   pre.right != current)
                pre = pre.right;
 
            if (pre.right == null)
            {
                 
                // Make current as the right
                // child of the right most node
                pre.right = current;
 
                // Traverse the left child
                current = current.left;
            }
            else
            {
                pre.right = null;
                succ = current;
                current = current.left;
                prev = null;
 
                // Traverse along the right
                // subtree to the
                // right-most child
                while (current != null)
                {
                    temp = current.right;
                    current.right = prev;
                    prev = current;
                    current = temp;
                }
 
                // Traverse back from
                // right most child to
                // current's left child node
                while (prev != null)
                {
                   Console.Write(prev.data + " ");
                    temp = prev.right;
                    prev.right = current;
                    current = prev;
                    prev = temp;
                }
                current = succ;
                current = current.right;
            }
        }
    }
}
 
// Driver code
static public void Main ()
{
     
    /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
      */
    PostOrder tree = new PostOrder();
    tree.root = new TreeNode(1);
    tree.root.left = new TreeNode(2);
    tree.root.right = new TreeNode(3);
    tree.root.left.left = new TreeNode(4);
    tree.root.left.right = new TreeNode(5);
    tree.root.right.left = new TreeNode(6);
    tree.root.right.right = new TreeNode(7);
    tree.root.left.right.left = new TreeNode(8);
    tree.root.left.right.right = new TreeNode(9);
 
    tree.postOrderConstantspace(tree.root);
}
}
 
// This code is contributed by offbeat


Javascript




<script>
 
// Javascript program to implement
// Post Order traversal
// of Binary Tree in O(N)
// time and O(1) space
 
// Definition of the
// binary tree
class TreeNode
{
 
    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }
}
 
 
var root;
 
// Function to find Post Order
// Traversal Using Constant space
function postOrderConstantspace(root)
{
    if (root == null)
        return;
   
    var current = new TreeNode(-1), pre = null;
    var prev = null,
             succ = null,
             temp = null;
              
    current.left = root;
 
    while (current != null)
    {
         
        // Go to the right child
        // if current does not
        // have a left child
        if (current.left == null)
        {
            current = current.right;
        }
 
        else
        {
             
            // Traverse left child
            pre = current.left;
 
            // Find the right most child
            // in the left subtree
            while (pre.right != null &&
                   pre.right != current)
                pre = pre.right;
 
            if (pre.right == null)
            {
                 
                // Make current as the right
                // child of the right most node
                pre.right = current;
 
                // Traverse the left child
                current = current.left;
            }
            else
            {
                pre.right = null;
                succ = current;
                current = current.left;
                prev = null;
 
                // Traverse along the right
                // subtree to the
                // right-most child
                while (current != null)
                {
                    temp = current.right;
                    current.right = prev;
                    prev = current;
                    current = temp;
                }
 
                // Traverse back from
                // right most child to
                // current's left child node
                while (prev != null)
                {
                   document.write(prev.data + " ");
                    temp = prev.right;
                    prev.right = current;
                    current = prev;
                    prev = temp;
                }
                current = succ;
                current = current.right;
            }
        }
    }
}
 
// Driver code
/* Constructed tree is as follows:-
                  1
               /     \
              2       3
             / \     / \
            4   5   6   7
               / \
              8   9
  */
var tree = new TreeNode();
tree.root = new TreeNode(1);
tree.root.left = new TreeNode(2);
tree.root.right = new TreeNode(3);
tree.root.left.left = new TreeNode(4);
tree.root.left.right = new TreeNode(5);
tree.root.right.left = new TreeNode(6);
tree.root.right.right = new TreeNode(7);
tree.root.left.right.left = new TreeNode(8);
tree.root.left.right.right = new TreeNode(9);
postOrderConstantspace(tree.root);
 
</script>


Output

4 8 9 5 2 6 7 3 1

Time Complexity: O(N) 
Auxiliary Space: O(1)

Method 2: In method 1, we traverse a path, reverse references, print nodes as we restore the references by reversing them again. In method 2, instead of reversing paths and restoring the structure, we traverse to the parent node from the current node using the current node’s left subtree. This could be faster depending on the tree structure, for example in a right-skewed tree. 

The following algorithm and diagrams provide the details of the approach.

Below is the conceptual diagram showing how the left and right child references are used to traverse back and forth. 

Below is the diagram which highlights the path 1->2->5->9 and the way the nodes are processed and printed as per the above algorithm. 

Below is the implementation of the above approach:

C++




// C++ Program to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
struct TreeNode {
    TreeNode* left;
    TreeNode* right;
    int data;
  
    TreeNode(int data)
    {
        this->data = data;
        this->left = nullptr;
        this->right = nullptr;
    }
};
 
TreeNode* root;
    
// Function to Calculate Post
// Order Traversal Using
// Constant Space
static void postOrderConstantspace(TreeNode* root)
{
  if (root == nullptr)
    return;
 
  TreeNode* current = nullptr;
  TreeNode* prevNode = nullptr;
  TreeNode* pre = nullptr;
  TreeNode* ptr = nullptr;
  TreeNode* netChild = nullptr;
  TreeNode* prevPtr = nullptr;
 
  current = root;
 
  while (current != nullptr)
  {
    if (current->left == nullptr)
    {
      current->left = prevNode;
 
      // Set prevNode to current
      prevNode = current;
      current = current->right;
    }
    else
    {
      pre = current->left;
 
      // Find the right most child
      // in the left subtree
      while (pre->right != nullptr &&
             pre->right != current)
        pre = pre->right;
 
      if (pre->right == nullptr)
      {
        pre->right = current;
        current = current->left;
      }
      else
      {
        // Set the right most
        // child's right pointer
        // to NULL
        pre->right = nullptr;
        cout << pre->data << " ";
        ptr = pre;
        netChild = pre;
        prevPtr = pre;
 
        while (ptr != nullptr)
        {
          if (ptr->right == netChild)
          {
            cout << ptr->data << " ";
            netChild = ptr;
            prevPtr->left = nullptr;
          }
 
          if (ptr == current->left)
            break;
 
          // Break the loop
          // all the left subtree
          // nodes of current
          // processed
          prevPtr = ptr;
          ptr = ptr->left;
        }
 
        prevNode = current;
        current = current->right;
      }
    }
  }
 
  cout << prevNode->data << " ";
 
  // Last path traversal
  // that includes the root.
  ptr = prevNode;
  netChild = prevNode;
  prevPtr = prevNode;
 
  while (ptr != nullptr)
  {
    if (ptr->right == netChild)
    {
      cout << ptr->data << " ";
      netChild = ptr;
      prevPtr->left = nullptr;
    }
    if (ptr == root)
      break;
 
    prevPtr = ptr;
    ptr = ptr->left;
  }
}
 
int main()
{
    /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
      */
  root = new TreeNode(1);
  root->left = new TreeNode(2);
  root->right = new TreeNode(3);
  root->left->left = new TreeNode(4);
  root->left->right = new TreeNode(5);
  root->right->left = new TreeNode(6);
  root->right->right = new TreeNode(7);
  root->left->right->left = new TreeNode(8);
  root->left->right->right = new TreeNode(9);
  postOrderConstantspace(root);
 
    return 0;
}
 
// This code is contributed by mukesh07.


Java




// Java Program to implement
// the above approach
class TreeNode {
    public int data;
    public TreeNode left;
    public TreeNode right;
 
    public TreeNode(int data)
    {
        this.data = data;
    }
 
    public String toString()
    {
        return data + " ";
    }
}
 
public class PostOrder {
    TreeNode root;
 
    // Function to Calculate Post
    // Order Traversal
    // Using Constant Space
    void postOrderConstantspace(TreeNode root)
    {
        if (root == null)
            return;
 
        TreeNode current = null;
        TreeNode prevNode = null;
        TreeNode pre = null;
        TreeNode ptr = null;
        TreeNode netChild = null;
        TreeNode prevPtr = null;
 
        current = root;
        while (current != null) {
            if (current.left == null) {
                current.left = prevNode;
                // Set prevNode to current
                prevNode = current;
                current = current.right;
            }
            else {
                pre = current.left;
                // Find the right most child
                // in the left subtree
                while (pre.right != null
                    && pre.right != current)
                    pre = pre.right;
 
                if (pre.right == null) {
                    pre.right = current;
                    current = current.left;
                }
                else {
                    // Set the right most
                    // child's right pointer
                    // to NULL
                    pre.right = null;
                    System.out.print(pre);
 
                    ptr = pre;
                    netChild = pre;
                    prevPtr = pre;
                    while (ptr != null) {
                        if (ptr.right == netChild) {
                            System.out.print(ptr);
                            netChild = ptr;
                            prevPtr.left = null;
                        }
 
                        if (ptr == current.left)
                            break;
                        // Break the loop
                        // all the left subtree
                        // nodes of current
                        // processed
 
                        prevPtr = ptr;
                        ptr = ptr.left;
                    }
 
                    prevNode = current;
                    current = current.right;
                }
            }
        }
 
        System.out.print(prevNode);
 
        // Last path traversal
        // that includes the root.
        ptr = prevNode;
        netChild = prevNode;
        prevPtr = prevNode;
        while (ptr != null) {
            if (ptr.right == netChild) {
                System.out.print(ptr);
                netChild = ptr;
                prevPtr.left = null;
            }
            if (ptr == root)
                break;
 
            prevPtr = ptr;
            ptr = ptr.left;
        }
    }
 
    // Main Function
    public static void main(String[] args)
    {
        /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
      */
        PostOrder tree = new PostOrder();
        tree.root = new TreeNode(1);
        tree.root.left = new TreeNode(2);
        tree.root.right = new TreeNode(3);
        tree.root.left.left
            = new TreeNode(4);
        tree.root.left.right
            = new TreeNode(5);
        tree.root.right.left
            = new TreeNode(6);
        tree.root.right.right
            = new TreeNode(7);
        tree.root.left.right.left
            = new TreeNode(8);
        tree.root.left.right.right
            = new TreeNode(9);
 
        tree.postOrderConstantspace(
            tree.root);
    }
}


Python3




# Python3 Program to implement the above approach
class TreeNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to Calculate Post
# Order Traversal Using
# Constant Space
def postOrderConstantspace(root):
  if root == None:
    return
 
  current = None
  prevNode = None
  pre = None
  ptr = None
  netChild = None
  prevPtr = None
 
  current = root
 
  while current != None:
    if current.left == None:
      current.left = prevNode
 
      # Set prevNode to current
      prevNode = current
      current = current.right
    else:
      pre = current.left
       
      # Find the right most child
      # in the left subtree
      while pre.right != None and pre.right != current:
        pre = pre.right
 
      if pre.right == None:
        pre.right = current
        current = current.left
      else:
        # Set the right most
        # child's right pointer
        # to NULL
        pre.right = None
        print(pre.data, end = " ")
        ptr = pre
        netChild = pre
        prevPtr = pre
 
        while ptr != None:
          if ptr.right == netChild:
            print(ptr.data, end = " ")
            netChild = ptr
            prevPtr.left = None
 
          if ptr == current.left:
            break
 
          # Break the loop
          # all the left subtree
          # nodes of current
          # processed
          prevPtr = ptr
          ptr = ptr.left
 
        prevNode = current
        current = current.right
 
  print(prevNode.data, end = " ")
 
  # Last path traversal
  # that includes the root.
  ptr = prevNode
  netChild = prevNode
  prevPtr = prevNode
 
  while ptr != None:
    if ptr.right == netChild:
      print(ptr.data, end = " ")
      netChild = ptr
      prevPtr.left = None
       
    if (ptr == root):
      break
 
    prevPtr = ptr
    ptr = ptr.left
 
""" Constructed tree is as follows:-
                  1
               /     \
              2       3
             / \     / \
            4   5   6   7
               / \
              8   9
"""
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)
root.left.right.left = TreeNode(8)
root.left.right.right = TreeNode(9)
postOrderConstantspace(root)
 
# This code is contributed by divyeshrabadiya07.


C#




// C# Program to implement
// the above approach
using System;
class TreeNode{
     
public int data;
public TreeNode left;
public TreeNode right;
  
public TreeNode(int data)
{
  this.data = data;
}
  
public string toString()
{
  return data + " ";
}
}
  
class PostOrder{
     
TreeNode root;
  
// Function to Calculate Post
// Order Traversal Using
// Constant Space
void postOrderConstantspace(TreeNode root)
{
  if (root == null)
    return;
 
  TreeNode current = null;
  TreeNode prevNode = null;
  TreeNode pre = null;
  TreeNode ptr = null;
  TreeNode netChild = null;
  TreeNode prevPtr = null;
 
  current = root;
   
  while (current != null)
  {
    if (current.left == null)
    {
      current.left = prevNode;
       
      // Set prevNode to current
      prevNode = current;
      current = current.right;
    }
    else
    {
      pre = current.left;
       
      // Find the right most child
      // in the left subtree
      while (pre.right != null &&
             pre.right != current)
        pre = pre.right;
 
      if (pre.right == null)
      {
        pre.right = current;
        current = current.left;
      }
      else
      {
        // Set the right most
        // child's right pointer
        // to NULL
        pre.right = null;
        Console.Write(pre.data + " ");
        ptr = pre;
        netChild = pre;
        prevPtr = pre;
         
        while (ptr != null)
        {
          if (ptr.right == netChild)
          {
            Console.Write(ptr.data + " ");
            netChild = ptr;
            prevPtr.left = null;
          }
 
          if (ptr == current.left)
            break;
           
          // Break the loop
          // all the left subtree
          // nodes of current
          // processed
          prevPtr = ptr;
          ptr = ptr.left;
        }
 
        prevNode = current;
        current = current.right;
      }
    }
  }
 
  Console.Write(prevNode.data + " ");
 
  // Last path traversal
  // that includes the root.
  ptr = prevNode;
  netChild = prevNode;
  prevPtr = prevNode;
   
  while (ptr != null)
  {
    if (ptr.right == netChild)
    {
      Console.Write(ptr.data + " ");
      netChild = ptr;
      prevPtr.left = null;
    }
    if (ptr == root)
      break;
 
    prevPtr = ptr;
    ptr = ptr.left;
  }
}
  
// Driver code
public static void Main(string[] args)
{
  /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
      */
  PostOrder tree = new PostOrder();
  tree.root = new TreeNode(1);
  tree.root.left = new TreeNode(2);
  tree.root.right = new TreeNode(3);
  tree.root.left.left = new TreeNode(4);
  tree.root.left.right = new TreeNode(5);
  tree.root.right.left = new TreeNode(6);
  tree.root.right.right = new TreeNode(7);
  tree.root.left.right.left = new TreeNode(8);
  tree.root.left.right.right = new TreeNode(9);
  tree.postOrderConstantspace(tree.root);
}
}
 
// This code is contributed by Rutvik_56


Javascript




<script>
    // Javascript Program to implement the above approach
     
    class TreeNode
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    let root;
   
    // Function to Calculate Post
    // Order Traversal Using
    // Constant Space
    function postOrderConstantspace(root)
    {
      if (root == null)
        return;
 
      let current = null;
      let prevNode = null;
      let pre = null;
      let ptr = null;
      let netChild = null;
      let prevPtr = null;
 
      current = root;
 
      while (current != null)
      {
        if (current.left == null)
        {
          current.left = prevNode;
 
          // Set prevNode to current
          prevNode = current;
          current = current.right;
        }
        else
        {
          pre = current.left;
 
          // Find the right most child
          // in the left subtree
          while (pre.right != null &&
                 pre.right != current)
            pre = pre.right;
 
          if (pre.right == null)
          {
            pre.right = current;
            current = current.left;
          }
          else
          {
            // Set the right most
            // child's right pointer
            // to NULL
            pre.right = null;
            document.write(pre.data + " ");
            ptr = pre;
            netChild = pre;
            prevPtr = pre;
 
            while (ptr != null)
            {
              if (ptr.right == netChild)
              {
                document.write(ptr.data + " ");
                netChild = ptr;
                prevPtr.left = null;
              }
 
              if (ptr == current.left)
                break;
 
              // Break the loop
              // all the left subtree
              // nodes of current
              // processed
              prevPtr = ptr;
              ptr = ptr.left;
            }
 
            prevNode = current;
            current = current.right;
          }
        }
      }
 
      document.write(prevNode.data + " ");
 
      // Last path traversal
      // that includes the root.
      ptr = prevNode;
      netChild = prevNode;
      prevPtr = prevNode;
 
      while (ptr != null)
      {
        if (ptr.right == netChild)
        {
          document.write(ptr.data + " ");
          netChild = ptr;
          prevPtr.left = null;
        }
        if (ptr == root)
          break;
 
        prevPtr = ptr;
        ptr = ptr.left;
      }
    }
     
    /* Constructed tree is as follows:-
                      1
                   /     \
                  2       3
                 / \     / \
                4   5   6   7
                   / \
                  8   9
    */
    root = new TreeNode(1);
    root.left = new TreeNode(2);
    root.right = new TreeNode(3);
    root.left.left = new TreeNode(4);
    root.left.right = new TreeNode(5);
    root.right.left = new TreeNode(6);
    root.right.right = new TreeNode(7);
    root.left.right.left = new TreeNode(8);
    root.left.right.right = new TreeNode(9);
    postOrderConstantspace(root);
     
    // This code is contributed by divyesh072019.
</script>


Output

4 8 9 5 2 6 7 3 1 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 


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