Morris traversal for Postorder

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Perform Postorder Tree traversal using Morris Traversal.

Examples:

Input: 1
/ \
2 3
/ \ / \
6 7 8 9

Output: 6 7 2 8 9 3 1

Input: 5
/ \
2 3
/ \ / \
4 N 8 9

Output: 4 2 8 9 3 5

Approach: The approach to performing Morris Traversal for Postorder is similar to Morris traversal for Preorder except swapping between left and right node links.

Create a vector and Initialize current as root
While current is not NULL
- If the current does not have a right child
  - push the current key in vector
    Go to the left, i.e., current = current->left
- else
  - Find leftmost node in current right subtree OR node whose left child == current.
- if current does not have a left child
  - push the current key in the vector
  - Make current as of the left child of that leftmost node
  - Go to this right child, i.e., current = current->right
- else
  - Found left child == current
- Update the left child as NULL of that node whose left child is current
- Go to the left, i.e. current = current->left
In the last, we reverse the vector and print it, since vector is used to store the output, the space complexity of this algorithm would be O(N).

Below is the implementation of the above approach

C++

// C++ program to perform
// Morris Traversal for Postorder
#include <bits/stdc++.h>
using namespace std;
 
struct TreeNode {
    int key;
    TreeNode* left;
    TreeNode* right;
 
    TreeNode(int data)
    {
        key = data;
        left = NULL;
        right = NULL;
    }
};
 
// Function to print vector
void print(vector<int>& ans)
{
    // Print the vector elements
    for (auto x : ans) {
        cout << x << " ";
    }
}
 
// Postorder traversal
// Without recursion and without stack
vector<int> postorderTraversal(TreeNode* root)
{
    vector<int> res;
    TreeNode* current = root;
 
    while (current != NULL) {
        // If right child is null,
        // put the current node data
        // in res. Move to left child.
        if (current->right == NULL) {
            res.push_back(current->key);
            current = current->left;
        }
        else {
            TreeNode* predecessor = current->right;
            while (predecessor->left != NULL
                   && predecessor->left != current) {
                predecessor = predecessor->left;
            }
            // If left child doesn't point
            // to this node, then put in res
            // this node and make left
            // child point to this node
            if (predecessor->left == NULL) {
                res.push_back(current->key);
                predecessor->left = current;
                current = current->right;
            }
            // If the left child of inorder predecessor
            // already points to this node
            else {
                predecessor->left = NULL;
                current = current->left;
            }
        }
    }
    // reverse the res
    reverse(res.begin(), res.end());
    return res;
}
// Driver program
int main()
{
    TreeNode* root = new TreeNode(10);
    root->left = new TreeNode(20);
    root->right = new TreeNode(30);
    root->right->left = new TreeNode(40);
    root->right->right = new TreeNode(50);
   
    cout << "Morris(postorder) Traversal: ";
    vector<int> ans = postorderTraversal(root);
   
    print(ans);
    return 0;
}

Java

// Java program to perform
// Morris Traversal for Postorder
import java.util.*;
class GFG{
 
static class TreeNode {
    int key;
    TreeNode left;
    TreeNode right;
 
    TreeNode(int data)
    {
        key = data;
        left = null;
        right = null;
    }
};
 
// Function to print vector
static void print(Vector<Integer> ans)
{
    // Print the vector elements
    for (int x : ans) {
        System.out.print(x+ " ");
    }
}
 
// Postorder traversal
// Without recursion and without stack
static Vector<Integer> postorderTraversal(TreeNode root)
{
    Vector<Integer> res = new Vector<>();
    TreeNode current = root;
 
    while (current != null)
    {
       
        // If right child is null,
        // put the current node data
        // in res. Move to left child.
        if (current.right == null) {
            res.add(current.key);
            current = current.left;
        }
        else {
            TreeNode predecessor = current.right;
            while (predecessor.left != null
                   && predecessor.left != current) {
                predecessor = predecessor.left;
            }
           
            // If left child doesn't point
            // to this node, then put in res
            // this node and make left
            // child point to this node
            if (predecessor.left == null) {
                res.add(current.key);
                predecessor.left = current;
                current = current.right;
            }
           
            // If the left child of inorder predecessor
            // already points to this node
            else {
                predecessor.left = null;
                current = current.left;
            }
        }
    }
   
    // reverse the res
    Collections.reverse(res);
    return res;
}
   
// Driver program
public static void main(String[] args)
{
    TreeNode root = new TreeNode(10);
    root.left = new TreeNode(20);
    root.right = new TreeNode(30);
    root.right.left = new TreeNode(40);
    root.right.right = new TreeNode(50);
   
    System.out.print("Morris(postorder) Traversal: ");
    Vector<Integer> ans = postorderTraversal(root);
   
    print(ans);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program  to perform
# Morris Traversal for Postorder
 
# class to create a new tree node
class TreeNode:
    def __init__(self, d):
        self.key = d
        self.left = None
        self.right = None
 
# Function to print list
def print1(ans) :
     
    # Print the vector elements
    for x in ans:
        print(x,end=" ")
     
# Postorder traversal
# Without recursion and without stack
def postorderTraversal(root) :
 
    res=[]
    current = root
 
    while current != None :
     
     
        # If right child is None,
        # put the current node data
        # in res. Move to left child.
        if current.right == None :
            res.append(current.key)
            current = current.left
         
        else :
            predecessor = current.right
            while predecessor.left != None and predecessor.left != current : 
                    predecessor = predecessor.left
             
             
            # If left child doesn't point
            # to this node, then put in res
            # this node and make left
            # child point to this node
            if predecessor.left == None:
                res.append(current.key)
                predecessor.left = current
                current = current.right
             
             
            # If the left child of inorder predecessor
            # already points to this node
            else :
                predecessor.left = None
                current = current.left
             
     
    # reverse the res
    res.reverse()
    return res
 
# Driver Code
if __name__ == '__main__':
     
    root = TreeNode(10)
    root.left = TreeNode(20)
    root.right = TreeNode(30)
    root.right.left = TreeNode(40)
    root.right.right = TreeNode(50)
     
    print("Morris(postorder) Traversal: ",end='')
    ans = postorderTraversal(root)
     
    print1(ans)
 
    # This code is contributed by jana_sayantan.

C#

// C# program to perform
// Morris Traversal for Postorder
using System;
using System.Collections.Generic;
 
public class GFG {
 
  public class TreeNode {
    public int key;
    public TreeNode left;
    public TreeNode right;
 
    public TreeNode(int data) {
      key = data;
      left = null;
      right = null;
    }
  };
 
  // Function to print vector
  static void print(List<int> ans) 
  {
     
    // Print the vector elements
    foreach (int x in ans) {
      Console.Write(x + " ");
    }
  }
 
  // Postorder traversal
  // Without recursion and without stack
  static List<int> postorderTraversal(TreeNode root) {
    List<int> res = new List<int>();
    TreeNode current = root;
 
    while (current != null) {
 
      // If right child is null,
      // put the current node data
      // in res. Move to left child.
      if (current.right == null) {
        res.Add(current.key);
        current = current.left;
      } else {
        TreeNode predecessor = current.right;
        while (predecessor.left != null && predecessor.left != current) {
          predecessor = predecessor.left;
        }
 
        // If left child doesn't point
        // to this node, then put in res
        // this node and make left
        // child point to this node
        if (predecessor.left == null) {
          res.Add(current.key);
          predecessor.left = current;
          current = current.right;
        }
 
        // If the left child of inorder predecessor
        // already points to this node
        else {
          predecessor.left = null;
          current = current.left;
        }
      }
    }
 
    // reverse the res
    res.Reverse();
    return res;
  }
 
  // Driver program
  public static void Main(String[] args) {
    TreeNode root = new TreeNode(10);
    root.left = new TreeNode(20);
    root.right = new TreeNode(30);
    root.right.left = new TreeNode(40);
    root.right.right = new TreeNode(50);
 
    Console.Write("Morris(postorder) Traversal: ");
    List<int> ans = postorderTraversal(root);
 
    print(ans);
  }
}
 
// This code is contributed by Rajput-Ji 

Javascript

<script>
 
// JavaScript program to perform
// Morris Traversal for Postorder
class TreeNode {
 
    constructor(data)
    {
        this.key = data;
        this.left = null;
        this.right = null;
    }
}
 
// Function to print vector
function print(ans)
{
 
    // Print the vector elements
    for (let x of ans) {
        document.write(x," ");
    }
}
 
// Postorder traversal
// Without recursion and without stack
function postorderTraversal(root)
{
    let res=[];
    let current = root;
 
    while (current != null) 
    {
     
        // If right child is null,
        // put the current node data
        // in res. Move to left child.
        if (current.right == null) {
            res.push(current.key);
            current = current.left;
        }
        else {
            let predecessor = current.right;
            while (predecessor.left != null
                   && predecessor.left != current) {
                predecessor = predecessor.left;
            }
             
            // If left child doesn't point
            // to this node, then put in res
            // this node and make left
            // child point to this node
            if (predecessor.left == null) {
                res.push(current.key);
                predecessor.left = current;
                current = current.right;
            }
             
            // If the left child of inorder predecessor
            // already points to this node
            else {
                predecessor.left = null;
                current = current.left;
            }
        }
    }
     
    // reverse the res
    res.reverse();
    return res;
}
 
// Driver program
 
let root = new TreeNode(10);
root.left = new TreeNode(20);
root.right = new TreeNode(30);
root.right.left = new TreeNode(40);
root.right.right = new TreeNode(50);
   
document.write("Morris(postorder) Traversal: ");
let ans = postorderTraversal(root);
   
print(ans);
 
// This code is contributed by shinjanpatra
 
</script>

Output

Morris(postorder) Traversal: 20 40 50 30 10

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

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