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 and a 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) return; if( k == 0 ) { cout << root->data << " "; return ; } else { 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 |
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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) return; if( k == 0 ) { printf( "%d ", root->data ); return ; } else { 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; } |
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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) return; if (k == 0) { System.out.print(node.data + " "); return; } else { 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 |
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Python
# Python program to find the nodes at k distance from root # A Binary tree node class Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None def 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) |
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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) { return; } if (k == 0) { Console.Write(node.data + " "); return; } else { 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 |
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Output:
4 5 8
Time Complexity: O(n) where n is number of nodes in the given binary tree.
Please write comments if you find the above code/algorithm incorrect, or find better ways to solve the same problem.
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