Find n-th node in Postorder traversal of a Binary Tree

Difficulty Level : Easy
Last Updated : 23 Jun, 2021

Given a Binary tree and a number N, write a program to find the N-th node in the Postorder traversal of the given Binary tree.
Prerequisite: Tree Traversal
Examples:

Input : N = 4
              11
            /   \
           21    31
         /   \
        41     51
Output : 31
Explanation: Postorder Traversal of given Binary Tree is 41 51 21 31 11, 
so 4th node will be 31.

Input : N = 5
             25
           /    \
          20    30
        /    \ /   \
      18    22 24   32
Output : 32

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The idea to solve this problem is to do postorder traversal of the given binary tree and keep track of the count of nodes visited while traversing the tree and print the current node when the count becomes equal to N.
Below is the implementation of the above approach:

C++

// C++ program to find n-th node of
// Postorder Traversal of Binary Tree
#include <bits/stdc++.h>
using namespace std;
 
// node of tree
struct Node {
    int data;
    Node *left, *right;
};
 
// function to create a new node
struct Node* createNode(int item)
{
    Node* temp = new Node;
    temp->data = item;
    temp->left = NULL;
    temp->right = NULL;
 
    return temp;
}
 
// function to find the N-th node in the postorder
// traversal of a given binary tree
void NthPostordernode(struct Node* root, int N)
{
    static int flag = 0;
 
    if (root == NULL)
        return;
 
    if (flag <= N) {
 
        // left recursion
        NthPostordernode(root->left, N);
 
        // right recursion
        NthPostordernode(root->right, N);
 
        flag++;
 
        // prints the n-th node of preorder traversal
        if (flag == N)
            cout << root->data;
    }
}
 
// driver code
int main()
{
    struct Node* root = createNode(25);
    root->left = createNode(20);
    root->right = createNode(30);
    root->left->left = createNode(18);
    root->left->right = createNode(22);
    root->right->left = createNode(24);
    root->right->right = createNode(32);
 
    int N = 6;
 
    // prints n-th node found
    NthPostordernode(root, N);
 
    return 0;
}

Java

// Java program to find n-th node of
// Postorder Traversal of Binary Tree
public class NthNodePostOrder {
 
    static int flag = 0;
 
    // function to find the N-th node in the postorder
    // traversal of a given binary tree
    public static void NthPostordernode(Node root, int N)
    {
   
        if (root == null)
            return;
   
        if (flag <= N)
        {  
            // left recursion
            NthPostordernode(root.left, N);
            // right recursion
            NthPostordernode(root.right, N);
            flag++;
            // prints the n-th node of preorder traversal
            if (flag == N)
                System.out.print(root.data);
        }
    }
 
 
    public static void main(String args[]) {
        Node root = new Node(25);
        root.left = new Node(20);
        root.right = new Node(30);
        root.left.left = new Node(18);
        root.left.right = new Node(22);
        root.right.left = new Node(24);
        root.right.right = new Node(32);
   
        int N = 6;
   
        // prints n-th node found
        NthPostordernode(root, N);
    }
}
 
/* A binary tree node structure */
class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data=data;
    }
};
// This code is contributed by Gaurav Tiwari

Python3

"""Python3 program to find n-th node of
Postorder Traversal of Binary Tree"""
 
# A Binary Tree Node
# Utility function to create a new tree node
class createNode:
 
    # Constructor to create a newNode
    def __init__(self, data):
        self.data= data
        self.left = None
        self.right = None
 
# function to find the N-th node
# in the postorder traversal of
# a given binary tree
flag = [0]
def NthPostordernode(root, N):
 
    if (root == None):
        return
 
    if (flag[0] <= N[0]):
         
        # left recursion
        NthPostordernode(root.left, N)
 
        # right recursion
        NthPostordernode(root.right, N)
 
        flag[0] += 1
 
        # prints the n-th node of
        # preorder traversal
        if (flag[0] == N[0]):
            print(root.data)
                         
# Driver Code
if __name__ == '__main__':
    root = createNode(25)
    root.left = createNode(20)
    root.right = createNode(30)
    root.left.left = createNode(18)
    root.left.right = createNode(22)
    root.right.left = createNode(24)
    root.right.right = createNode(32)
 
    N = [6]
 
    # prints n-th node found
    NthPostordernode(root, N)
 
# This code is contributed by
# SHUBHAMSINGH10

C#

// C# program to find n-th node of
// Postorder Traversal of Binary Tree
using System;
 
public class NthNodePostOrder
{
 
    /* A binary tree node structure */
    public class Node
    {
        public int data;
        public Node left, right;
        public Node(int data)
        {
            this.data=data;
        }
    }
    static int flag = 0;
 
    // function to find the N-th node in the postorder
    // traversal of a given binary tree
    static void NthPostordernode(Node root, int N)
    {
        if (root == null)
            return;
     
        if (flag <= N)
        {
            // left recursion
            NthPostordernode(root.left, N);
             
            // right recursion
            NthPostordernode(root.right, N);
            flag++;
             
            // prints the n-th node of preorder traversal
            if (flag == N)
                Console.Write(root.data);
        }
    }
 
    // Driver code
    public static void Main(String []args)
    {
        Node root = new Node(25);
        root.left = new Node(20);
        root.right = new Node(30);
        root.left.left = new Node(18);
        root.left.right = new Node(22);
        root.right.left = new Node(24);
        root.right.right = new Node(32);
     
        int N = 6;
     
        // prints n-th node found
        NthPostordernode(root, N);
    }
}
 
// This code is contributed by Arnab Kundu

Javascript

<script>
 
// JavaScript program to find n-th node of
// Postorder Traversal of Binary Tree
/* A binary tree node structure */
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}
 
var flag = 0;
// function to find the N-th node in the postorder
// traversal of a given binary tree
function NthPostordernode(root, N)
{
    if (root == null)
        return;
 
    if (flag <= N)
    {
        // left recursion
        NthPostordernode(root.left, N);
         
        // right recursion
        NthPostordernode(root.right, N);
        flag++;
         
        // prints the n-th node of preorder traversal
        if (flag == N)
            document.write(root.data);
    }
}
 
// Driver code
var root = new Node(25);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(18);
root.left.right = new Node(22);
root.right.left = new Node(24);
root.right.right = new Node(32);
var N = 6;
// prints n-th node found
NthPostordernode(root, N);
 
</script>

Output:

Time Complexity: O(n), where n is the number of nodes in the given binary tree.
Auxiliary Space: O(1)

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