Print all nodes that are at distance k from a leaf node

Given a Binary Tree and a positive integer k, print all nodes that are distance k from a leaf node.

Here the meaning of distance is different from previous post. Here k distance from a leaf means k levels higher than a leaf node. For example if k is more than height of Binary Tree, then nothing should be printed. Expected time complexity is O(n) where n is the number nodes in the given Binary Tree.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to traverse the tree. Keep storing all ancestors till we hit a leaf node. When we reach a leaf node, we print the ancestor at distance k. We also need to keep track of nodes that are already printed as output. For that we use a boolean array visited[].

C++

/* Program to print all nodes which are at distance k from a leaf */
#include <iostream> 
using namespace std; 
#define MAX_HEIGHT 10000 
  
struct Node 
{ 
    int key; 
    Node *left, *right; 
}; 
  
/* utility that allocates a new Node with the given key  */
Node* newNode(int key) 
{ 
    Node* node = new Node; 
    node->key = key; 
    node->left = node->right = NULL; 
    return (node); 
} 
  
/* This function prints all nodes that are distance k from a leaf node 
   path[] --> Store ancestors of a node 
   visited[] --> Stores true if a node is printed as output.  A node may be k 
                 distance away from many leaves, we want to print it once */
void kDistantFromLeafUtil(Node* node, int path[], bool visited[], 
                          int pathLen, int k) 
{ 
    // Base case 
    if (node==NULL) return; 
  
    /* append this Node to the path array */
    path[pathLen] = node->key; 
    visited[pathLen] = false; 
    pathLen++; 
  
    /* it's a leaf, so print the ancestor at distance k only 
       if the ancestor is not already printed  */
    if (node->left == NULL && node->right == NULL && 
        pathLen-k-1 >= 0 && visited[pathLen-k-1] == false) 
    { 
        cout << path[pathLen-k-1] << " "; 
        visited[pathLen-k-1] = true; 
        return; 
    } 
  
    /* If not leaf node, recur for left and right subtrees */
    kDistantFromLeafUtil(node->left, path, visited, pathLen, k); 
    kDistantFromLeafUtil(node->right, path, visited, pathLen, k); 
} 
  
/* Given a binary tree and a nuber k, print all nodes that are k 
   distant from a leaf*/
void printKDistantfromLeaf(Node* node, int k) 
{ 
    int path[MAX_HEIGHT]; 
    bool visited[MAX_HEIGHT] = {false}; 
    kDistantFromLeafUtil(node, path, visited, 0, k); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    // Let us create binary tree given in the above example 
    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(6); 
    root->right->right = newNode(7); 
    root->right->left->right = newNode(8); 
  
    cout << "Nodes at distance 2 are: "; 
    printKDistantfromLeaf(root, 2); 
  
    return 0; 
} 

Java

// Java program to print all nodes at a distance k from leaf 
// A binary tree node 
class Node 
{ 
    int data; 
    Node left, right; 
   
    Node(int item) 
    { 
        data = item; 
        left = right = null; 
    } 
} 
   
class BinaryTree 
{ 
    Node root; 
   
    /* This function prints all nodes that are distance k from a leaf node 
     path[] --> Store ancestors of a node 
     visited[] --> Stores true if a node is printed as output.  A node may 
     be k distance away from many leaves, we want to print it once */
    void kDistantFromLeafUtil(Node node, int path[], boolean visited[], 
                              int pathLen, int k) 
    { 
        // Base case 
        if (node == null) 
            return; 
   
        /* append this Node to the path array */
        path[pathLen] = node.data; 
        visited[pathLen] = false; 
        pathLen++; 
   
        /* it's a leaf, so print the ancestor at distance k only 
         if the ancestor is not already printed  */
        if (node.left == null && node.right == null
           && pathLen - k - 1 >= 0 && visited[pathLen - k - 1] == false) 
        { 
            System.out.print(path[pathLen - k - 1] + " "); 
            visited[pathLen - k - 1] = true; 
            return; 
        } 
   
        /* If not leaf node, recur for left and right subtrees */
        kDistantFromLeafUtil(node.left, path, visited, pathLen, k); 
        kDistantFromLeafUtil(node.right, path, visited, pathLen, k); 
    } 
   
    /* Given a binary tree and a nuber k, print all nodes that are k 
     distant from a leaf*/
    void printKDistantfromLeaf(Node node, int k) 
    { 
        int path[] = new int[1000]; 
        boolean visited[] = new boolean[1000]; 
        kDistantFromLeafUtil(node, path, visited, 0, k); 
    } 
   
    // Driver program to test the above functions 
    public static void main(String args[]) 
    { 
        BinaryTree tree = new BinaryTree(); 
   
        /* Let us construct the tree shown in above diagram */
        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(6); 
        tree.root.right.right = new Node(7); 
        tree.root.right.left.right = new Node(8); 
   
        System.out.println(" Nodes at distance 2 are :"); 
        tree.printKDistantfromLeaf(tree.root, 2); 
    } 
} 
   
// This code has been contributed by Mayank Jaiswal 

Python3

# Program to print all nodes which are at  
# distance k from a leaf  
  
# utility that allocates a new Node with  
# the given key  
class newNode: 
    def __init__(self, key): 
        self.key = key  
        self.left = self.right = None
  
# This function prints all nodes that  
# are distance k from a leaf node  
# path[] -. Store ancestors of a node  
# visited[] -. Stores true if a node is  
# printed as output. A node may be k distance  
# away from many leaves, we want to print it once  
def kDistantFromLeafUtil(node, path, visited,  
                                 pathLen, k): 
      
    # Base case  
    if (node == None): 
        return
  
    # append this Node to the path array 
    path[pathLen] = node.key  
    visited[pathLen] = False
    pathLen += 1
  
    # it's a leaf, so print the ancestor at  
    # distance k only if the ancestor is  
    # not already printed  
    if (node.left == None and node.right == None and 
                            pathLen - k - 1 >= 0 and 
                            visited[pathLen - k - 1] == False): 
        print(path[pathLen - k - 1], end = " ")  
        visited[pathLen - k - 1] = True
        return
  
    # If not leaf node, recur for left  
    # and right subtrees  
    kDistantFromLeafUtil(node.left, path,  
                         visited, pathLen, k) 
    kDistantFromLeafUtil(node.right, path,  
                         visited, pathLen, k) 
  
# Given a binary tree and a nuber k,  
# print all nodes that are k distant from a leaf 
def printKDistantfromLeaf(node, k): 
    global MAX_HEIGHT 
    path = [None] * MAX_HEIGHT  
    visited = [False] * MAX_HEIGHT 
    kDistantFromLeafUtil(node, path, visited, 0, k) 
  
# Driver Code 
MAX_HEIGHT = 10000
  
# Let us create binary tree given in  
# the above example  
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(6)  
root.right.right = newNode(7)  
root.right.left.right = newNode(8)  
  
print("Nodes at distance 2 are:", end = " ")  
printKDistantfromLeaf(root, 2) 
  
# This code is contributed by pranchalK 

C#

using System; 
  
// C# program to print all nodes at a distance k from leaf  
// A binary tree node  
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; 
  
    /* This function prints all nodes that are distance k from a leaf node  
     path[] --> Store ancestors of a node  
     visited[] --> Stores true if a node is printed as output.  A node may  
     be k distance away from many leaves, we want to print it once */
    public virtual void kDistantFromLeafUtil(Node node, int[] path, bool[] visited, int pathLen, int k) 
    { 
        // Base case  
        if (node == null) 
        { 
            return; 
        } 
  
        /* append this Node to the path array */
        path[pathLen] = node.data; 
        visited[pathLen] = false; 
        pathLen++; 
  
        /* it's a leaf, so print the ancestor at distance k only  
         if the ancestor is not already printed  */
        if (node.left == null && node.right == null && pathLen - k - 1 >= 0 && visited[pathLen - k - 1] == false) 
        { 
            Console.Write(path[pathLen - k - 1] + " "); 
            visited[pathLen - k - 1] = true; 
            return; 
        } 
  
        /* If not leaf node, recur for left and right subtrees */
        kDistantFromLeafUtil(node.left, path, visited, pathLen, k); 
        kDistantFromLeafUtil(node.right, path, visited, pathLen, k); 
    } 
  
    /* Given a binary tree and a nuber k, print all nodes that are k  
     distant from a leaf*/
    public virtual void printKDistantfromLeaf(Node node, int k) 
    { 
        int[] path = new int[1000]; 
        bool[] visited = new bool[1000]; 
        kDistantFromLeafUtil(node, path, visited, 0, k); 
    } 
  
    // Driver program to test the above functions  
    public static void Main(string[] args) 
    { 
        BinaryTree tree = new BinaryTree(); 
  
        /* Let us construct the tree shown in above diagram */
        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(6); 
        tree.root.right.right = new Node(7); 
        tree.root.right.left.right = new Node(8); 
  
        Console.WriteLine(" Nodes at distance 2 are :"); 
        tree.printKDistantfromLeaf(tree.root, 2); 
    } 
} 
  
// This code is contributed by Shrikant13 

Output:

Nodes at distance 2 are: 1 3

Time Complexity: Time Complexity of above code is O(n) as the code does a simple tree traversal.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

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