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Postorder successor of a Node in Binary Tree
  • Difficulty Level : Medium
  • Last Updated : 04 Jun, 2019

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 :  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.

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

An efficient solution is based on below observations.

  1. If given node is root then postorder successor is NULL, since root is the last node print in a postorder traversal
  2. If given node is right child of parent or right child of parent is NULL, then parent is postorder successor.
  3. 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

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 substree 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 substree 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 substree 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 substree 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


Output:

Postorder successor of 19 is 18

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

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