Data Modelling & AI Data Structure & Algorithm

Postorder successor of a Node in Binary Tree

21 June 2025

2

Given a binary tree and a node in the binary tree, find Postorder successor of the given node.

Examples: Consider the following binary tree
              20            
           /      \         
          10       26       
         /  \     /   \     
       4     18  24    27   
            /  \
           14   19
          /  \
         13  15

Postorder traversal of given tree is 4, 13, 15, 14,
19, 18, 10, 24, 27, 26, 20.

Input :  Complexity Analysis:24
Output : 27

Input : 4
Output : 13

A simple solution is to first store Postorder traversal of the given tree in an array then linearly search given node and print node next to it.

Complexity Analysis:

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

An efficient solution is based on below observations.

If given node is root then postorder successor is NULL, since root is the last node print in a postorder traversal
If given node is right child of parent or right child of parent is NULL, then parent is postorder successor.
If given node is left child of parent and right child of parent is not NULL, then postorder successor is the leftmost node of parent’s right subtree

Implementation:

C++

// CPP program to find postorder successor of
// given node.
#include <iostream>
using namespace std;
  
struct Node {
    struct Node *left, *right, *parent;
    int value;
};
  
// Utility function to create a new node with
// given value.
struct Node* newNode(int value)
{
    Node* temp = new Node;
    temp->left = temp->right = temp->parent = NULL;
    temp->value = value;
    return temp;
}
  
Node* postorderSuccessor(Node* root, Node* n)
{
    // Root has no successor in postorder
    // traversal
    if (n == root)
        return NULL;
  
    // If given node is right child of its
    // parent or parent's right is empty, then 
    // parent is postorder successor.
    Node* parent = n->parent;
    if (parent->right == NULL || parent->right == n)
        return parent;
  
    // In all other cases, find the leftmost 
    // child in right subtree of parent.
    Node* curr = parent->right;
    while (curr->left != NULL)
        curr = curr->left;
  
    return curr;
}
  
// Driver code
int main()
{
    struct Node* root = newNode(20);
    root->parent = NULL;
    root->left = newNode(10);
    root->left->parent = root;
    root->left->left = newNode(4);
    root->left->left->parent = root->left;
    root->left->right = newNode(18);
    root->left->right->parent = root->left;
    root->right = newNode(26);
    root->right->parent = root;
    root->right->left = newNode(24);
    root->right->left->parent = root->right;
    root->right->right = newNode(27);
    root->right->right->parent = root->right;
    root->left->right->left = newNode(14);
    root->left->right->left->parent = root->left->right;
    root->left->right->left->left = newNode(13);
    root->left->right->left->left->parent = root->left->right->left;
    root->left->right->left->right = newNode(15);
    root->left->right->left->right->parent = root->left->right->left;
    root->left->right->right = newNode(19);
    root->left->right->right->parent = root->left->right;
  
    struct Node* res = postorderSuccessor(root, root->left->right->right);
    if (res) 
        printf("Postorder successor of %d is %d\n",
               root->left->right->right->value, res->value);    
    else
        printf("Postorder successor of %d is NULL\n",
               root->left->right->right->value);    
  
    return 0;
}

Java

// Java program to find postorder successor of 
// given node. 
import java.util.*;
class GfG {
 
static class Node { 
    Node left, right, parent; 
    int value; 
}
 
// Utility function to create a new node with 
// given value. 
static Node newNode(int value) 
{ 
    Node temp = new Node(); 
    temp.left = null;
    temp.right = null;
    temp.parent = null; 
    temp.value = value; 
    return temp; 
} 
 
static Node postorderSuccessor(Node root, Node n) 
{ 
    // Root has no successor in postorder 
    // traversal 
    if (n == root) 
        return null; 
 
    // If given node is right child of its 
    // parent or parent's right is empty, then 
    // parent is postorder successor. 
    Node parent = n.parent; 
    if (parent.right == null || parent.right == n) 
        return parent; 
 
    // In all other cases, find the leftmost 
    // child in right subtree of parent. 
    Node curr = parent.right; 
    while (curr.left != null) 
        curr = curr.left; 
 
    return curr; 
} 
 
// Driver code 
public static void main(String[] args) 
{ 
    Node root = newNode(20); 
    root.parent = null; 
    root.left = newNode(10); 
    root.left.parent = root; 
    root.left.left = newNode(4); 
    root.left.left.parent = root.left; 
    root.left.right = newNode(18); 
    root.left.right.parent = root.left; 
    root.right = newNode(26); 
    root.right.parent = root; 
    root.right.left = newNode(24); 
    root.right.left.parent = root.right; 
    root.right.right = newNode(27); 
    root.right.right.parent = root.right; 
    root.left.right.left = newNode(14); 
    root.left.right.left.parent = root.left.right; 
    root.left.right.left.left = newNode(13); 
    root.left.right.left.left.parent = root.left.right.left; 
    root.left.right.left.right = newNode(15); 
    root.left.right.left.right.parent = root.left.right.left; 
    root.left.right.right = newNode(19); 
    root.left.right.right.parent = root.left.right; 
 
    Node res = postorderSuccessor(root, root.left.right.right); 
    if (res != null) 
        System.out.println("Postorder successor of "+ 
        root.left.right.right.value + " is "+ res.value); 
    else
        System.out.println("Postorder successor of " + 
        root.left.right.right.value + " is NULL");
} 
}

Python3

""" Python3 program to find postorder 
successor of a node in Binary Tree."""
 
# A Binary Tree Node 
# Utility function to create a new tree node 
class newNode: 
 
    # Constructor to create a new node 
    def __init__(self, data): 
        self.value = data 
        self.left = None
        self.right = None
        self.parent=None
 
def postorderSuccessor(root, n) :
 
    # Root has no successor in postorder 
    # traversal 
    if (n == root):
        return None
     
    # If given node is right child of its 
    # parent or parent's right is empty, 
    # then parent is postorder successor. 
    parent = n.parent 
    if (parent.right == None or parent.right == n):
        return parent 
     
    # In all other cases, find the leftmost 
    # child in right subtree of parent. 
    curr = parent.right 
    while (curr.left != None):
        curr = curr.left 
     
    return curr
     
# Driver Code
if __name__ == '__main__':
    root = newNode(20) 
    root.parent = None
    root.left = newNode(10) 
    root.left.parent = root 
    root.left.left = newNode(4) 
    root.left.left.parent = root.left 
    root.left.right = newNode(18) 
    root.left.right.parent = root.left 
    root.right = newNode(26) 
    root.right.parent = root 
    root.right.left = newNode(24) 
    root.right.left.parent = root.right 
    root.right.right = newNode(27) 
    root.right.right.parent = root.right 
    root.left.right.left = newNode(14) 
    root.left.right.left.parent = root.left.right 
    root.left.right.left.left = newNode(13) 
    root.left.right.left.left.parent = root.left.right.left 
    root.left.right.left.right = newNode(15) 
    root.left.right.left.right.parent = root.left.right.left 
    root.left.right.right = newNode(19) 
    root.left.right.right.parent = root.left.right
    res = postorderSuccessor(root, root.left.right.right) 
 
    if (res) : 
        print("postorder successor of", 
               root.left.right.right.value, 
               "is", res.value) 
     
    else:
        print("postorder successor of", 
               root.left.right.right.value, 
                                 "is None")
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# program to find postorder successor of 
// given node.
using System;
 
class GfG 
{
 
class Node 
{ 
    public Node left, right, parent; 
    public int value; 
}
 
// Utility function to create  
// a new node with given value. 
static Node newNode(int value) 
{ 
    Node temp = new Node(); 
    temp.left = null;
    temp.right = null;
    temp.parent = null; 
    temp.value = value; 
    return temp; 
} 
 
static Node postorderSuccessor(Node root, Node n) 
{ 
    // Root has no successor in  
    // postorder traversal 
    if (n == root) 
        return null; 
 
    // If given node is right child of its 
    // parent or parent's right is empty, then 
    // parent is postorder successor. 
    Node parent = n.parent; 
    if (parent.right == null || parent.right == n) 
        return parent; 
 
    // In all other cases, find the leftmost 
    // child in right subtree of parent. 
    Node curr = parent.right; 
    while (curr.left != null) 
        curr = curr.left; 
 
    return curr; 
} 
 
// Driver code 
public static void Main(String[] args) 
{ 
    Node root = newNode(20); 
    root.parent = null; 
    root.left = newNode(10); 
    root.left.parent = root; 
    root.left.left = newNode(4); 
    root.left.left.parent = root.left; 
    root.left.right = newNode(18); 
    root.left.right.parent = root.left; 
    root.right = newNode(26); 
    root.right.parent = root; 
    root.right.left = newNode(24); 
    root.right.left.parent = root.right; 
    root.right.right = newNode(27); 
    root.right.right.parent = root.right; 
    root.left.right.left = newNode(14); 
    root.left.right.left.parent = root.left.right; 
    root.left.right.left.left = newNode(13); 
    root.left.right.left.left.parent = root.left.right.left; 
    root.left.right.left.right = newNode(15); 
    root.left.right.left.right.parent = root.left.right.left; 
    root.left.right.right = newNode(19); 
    root.left.right.right.parent = root.left.right; 
 
    Node res = postorderSuccessor(root, root.left.right.right); 
    if (res != null) 
        Console.WriteLine("Postorder successor of "+ 
        root.left.right.right.value + " is "+ res.value); 
    else
        Console.WriteLine("Postorder successor of " + 
        root.left.right.right.value + " is NULL");
} 
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// javascript program to find postorder successor of 
// given node.
 
class Node 
{ 
    constructor()
    {
        this.value = 0;
        this.left = null;
        this.right = null;
        this.parent = null;
    }
}
 
// Utility function to create  
// a new node with given value. 
function newNode(value) 
{ 
    var temp = new Node(); 
    temp.left = null;
    temp.right = null;
    temp.parent = null; 
    temp.value = value; 
    return temp; 
} 
 
function postorderSuccessor(root, n) 
{ 
    // Root has no successor in  
    // postorder traversal 
    if (n == root) 
        return null; 
 
    // If given node is right child of its 
    // parent or parent's right is empty, then 
    // parent is postorder successor. 
    var parent = n.parent; 
    if (parent.right == null || parent.right == n) 
        return parent; 
 
    // In all other cases, find the leftmost 
    // child in right subtree of parent. 
    var curr = parent.right; 
    while (curr.left != null) 
        curr = curr.left; 
 
    return curr; 
} 
 
// Driver code 
var root = newNode(20); 
root.parent = null; 
root.left = newNode(10); 
root.left.parent = root; 
root.left.left = newNode(4); 
root.left.left.parent = root.left; 
root.left.right = newNode(18); 
root.left.right.parent = root.left; 
root.right = newNode(26); 
root.right.parent = root; 
root.right.left = newNode(24); 
root.right.left.parent = root.right; 
root.right.right = newNode(27); 
root.right.right.parent = root.right; 
root.left.right.left = newNode(14); 
root.left.right.left.parent = root.left.right; 
root.left.right.left.left = newNode(13); 
root.left.right.left.left.parent = root.left.right.left; 
root.left.right.left.right = newNode(15); 
root.left.right.left.right.parent = root.left.right.left; 
root.left.right.right = newNode(19); 
root.left.right.right.parent = root.left.right; 
var res = postorderSuccessor(root, root.left.right.right); 
if (res != null) 
    document.write("Postorder successor of "+ 
    root.left.right.right.value + " is "+ res.value); 
else
    document.write("Postorder successor of " + 
    root.left.right.right.value + " is NULL");
 
// This code is contributed by famously.
 
</script>

Output

Postorder successor of 19 is 18

Complexity Analysis:

Time Complexity : O(h) where h is height of given Binary Tree.
Auxiliary Space : O(1) since no use of arrays, stacks, queues.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

2 COMMENTS

lsm99k 26 January 2026 At 7:42 am

… [Trackback]

[…] Information to that Topic: geeksforgeeks.org/postorder-successor-of-a-node-in-binary-tree/ […]

Log in to leave a comment
สล็อตเว็บตรง 10 March 2026 At 10:51 am

… [Trackback]

[…] Read More on that Topic: geeksforgeeks.org/postorder-successor-of-a-node-in-binary-tree/ […]

Log in to leave a comment

Postorder successor of a Node in Binary Tree

C++

Java

Python3

C#

Javascript

Adding Persistent Memory to Claude Code with the Lightweight memsearch Plugin

GLM-5 vs. MiniMax M2.5 vs. Gemini 3 Deep Think: Which Model Fits Your AI Agent Stack?

We Extracted OpenClaw’s Memory System and Open-Sourced It (memsearch)

2 COMMENTS

LEAVE A REPLY Cancel reply

Most Popular

Samsung’s June update brings Galaxy S25 three powerful AI tricks the S26 already had

Your Galaxy S25 just picked up three more Galaxy AI features

No one can agree on these phones. [Video]

How to Buy & Sell NFTs (And Stay 100% Safe) in 2026 by Katarina Glamoslija

EDITOR PICKS

Samsung’s June update brings Galaxy S25 three powerful AI tricks the S26 already had

Your Galaxy S25 just picked up three more Galaxy AI features

No one can agree on these phones. [Video]

POPULAR POSTS

Samsung’s June update brings Galaxy S25 three powerful AI tricks the S26 already had

Your Galaxy S25 just picked up three more Galaxy AI features

No one can agree on these phones. [Video]

POPULAR CATEGORY

ABOUT US

FOLLOW US