Print nodes at k distance from root
Given a root of a tree, and an integer k. Print all the nodes which are at k distance from root.
For example, in the below tree, 4, 5 & 8 are at distance 2 from root.
1
/ \
2 3
/ \ /
4 5 8
The problem can be solved using recursion. Thanks to eldho for suggesting the solution.
C++
#include<bits/stdc++.h>using namespace std;/* A binary tree node has data,pointer to left child anda pointer to right child */class node{ public: int data; node* left; node* right; /* Constructor that allocates a new node with the given data and NULL left and right pointers. */ node(int data) { this->data = data; this->left = NULL; this->right = NULL; }};void printKDistant(node *root , int k){ if(root == NULL|| k < 0 ) return; if( k == 0 ) { cout << root->data << " "; return; } printKDistant( root->left, k - 1 ) ; printKDistant( root->right, k - 1 ) ; }/* Driver code*/int main(){ /* Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8 */ node *root = new node(1); root->left = new node(2); root->right = new node(3); root->left->left = new node(4); root->left->right = new node(5); root->right->left = new node(8); printKDistant(root, 2); return 0;}// This code is contributed by rathbhupendra |
C
#include <stdio.h>#include <stdlib.h>/* A binary tree node has data, pointer to left child and a pointer to right child */struct node{ int data; struct node* left; struct node* right;};void printKDistant(struct node *root , int k) { if(root == NULL|| k < 0 ) return; if( k == 0 ) { printf( "%d ", root->data ); return ; } printKDistant( root->left, k-1 ) ; printKDistant( root->right, k-1 ) ; }/* Helper function that allocates a new node with the given data and NULL left and right pointers. */struct node* newNode(int data){ struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node);}/* Driver program to test above functions*/int main(){ /* Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8 */ struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(8); printKDistant(root, 2); getchar(); return 0;} |
Java
// Java program to print nodes at k distance from root /* A binary tree node has data, pointer to left child and a pointer to right child */class Node{ int data; Node left, right; Node(int item) { data = item; left = right = null; }} class BinaryTree{ Node root; void printKDistant(Node node, int k) { if (node == null|| k < 0 ) //Base case return; if (k == 0) { System.out.print(node.data + " "); return; } //recursively traversing printKDistant(node.left, k - 1); printKDistant(node.right, k - 1); } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); /* Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8 */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(8); tree.printKDistant(tree.root, 2); }} // This code has been contributed by Mayank Jaiswal |
Python
# Python program to find the nodes at k distance from root# A Binary tree nodeclass Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = Nonedef printKDistant(root, k): if root is None: return if k == 0: print root.data, else: printKDistant(root.left, k-1) printKDistant(root.right, k-1)# Driver program to test above function""" Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8"""root = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.left.right = Node(5)root.right.left = Node(8)printKDistant(root, 2)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// c# program to print nodes at k distance from root/* A binary tree node has data, pointer to left child and a pointer to right child */public class Node{ public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree{ public Node root; public virtual void printKDistant(Node node, int k) { if (node == null|| k < 0 ) { return; } if (k == 0) { Console.Write(node.data + " "); return; } printKDistant(node.left, k - 1); printKDistant(node.right, k - 1); } /* Driver program to test above functions */ public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); /* Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8 */ tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(8); tree.printKDistant(tree.root, 2); }}// This code is contributed by Shrikant13 |
Javascript
<script>// Javascript program to print nodes at k distance from root/* A binary tree node has data, pointer to left child and a pointer to right child */class Node{ constructor(item) { this.data = item; this.left = null; this.right = null; }}var root =null;function printKDistant(node, k){ if (node == null|| k < 0 ) { return; } if (k == 0) { document.write(node.data + " "); return; } printKDistant(node.left, k - 1); printKDistant(node.right, k - 1); }/* Driver program to test above functions *//* Constructed binary tree is 1 / \ 2 3 / \ / 4 5 8 */root = new Node(1);root.left = new Node(2);root.right = new Node(3);root.left.left = new Node(4);root.left.right = new Node(5);root.right.left = new Node(8);printKDistant(root, 2);// This code is contributed by importantly.</script> |
Output:
4 5 8
Time Complexity: O(n) where n is number of nodes in the given binary tree.
Space Complexity : O(height of the binary tree).
Note-
- If it’s true print the node – Always check the K distance == 0 at every node
- the left or right subtree – Decrement the distance by 1 when you are passing to its subtree
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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