Find n-th node in Postorder traversal of a Binary Tree
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
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 |
Output:
30
Time Complexity: O(n), where n is the number of nodes in the given binary tree.
Auxiliary Space: O(1)
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