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.
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 nodeswhich are at distance k from a leaf */#include <iostream>using namespace std;#define MAX_HEIGHT 10000struct 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 number 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 code*/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 nodeclass 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 number 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 keyclass 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 oncedef 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 number k,# print all nodes that are k distant from a leafdef printKDistantfromLeaf(node, k): global MAX_HEIGHT path = [None] * MAX_HEIGHT visited = [False] * MAX_HEIGHT kDistantFromLeafUtil(node, path, visited, 0, k)# Driver CodeMAX_HEIGHT = 10000# Let us create binary tree given in# the above exampleroot = 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 nodepublic 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 number 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 |
Javascript
<script> // JavaScript program to print all // nodes at a distance k from leaf // A binary tree node class Node { constructor(item) { this.left = null; this.right = null; this.data = item; } } let 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 */ function kDistantFromLeafUtil(node, path, visited, pathLen, 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) { document.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 number k, print all nodes that are k distant from a leaf*/ function printKDistantfromLeaf(node, k) { let path = new Array(1000); path.fill(0); let visited = new Array(1000); visited.fill(false); kDistantFromLeafUtil(node, path, visited, 0, k); } /* Let us construct the tree shown in above diagram */ 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(6); root.right.right = new Node(7); root.right.left.right = new Node(8); document.write(" Nodes at distance 2 are : "); printKDistantfromLeaf(root, 2); </script> |
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.
Space Optimized Solution :
C++
// C++ program to print all nodes at a distance k from leaf// A binary tree node#include <bits/stdc++.h>using namespace std;struct Node{ int data; Node *left, *right;};// Utility function to // create a new tree node Node* newNode(int key) { Node *temp = new Node; temp->data= key; temp->left = temp->right = NULL; return temp; }/* Given a binary tree and a number k,print all nodes that are kdistant from a leaf*/int printKDistantfromLeaf(struct Node *node, int k){ if (node == NULL) return -1; int lk = printKDistantfromLeaf(node->left, k); int rk = printKDistantfromLeaf(node->right, k); bool isLeaf = lk == -1 && lk == rk; if (lk == 0 || rk == 0 || (isLeaf && k == 0)) cout<<(" " )<<( node->data); if (isLeaf && k > 0) return k - 1; // leaf node if (lk > 0 && lk < k) return lk - 1; // parent of left leaf if (rk > 0 && rk < k) return rk - 1; // parent of right leaf return -2;}// Driver codeint main(){ Node *root = NULL; /* Let us construct the tree shown in above diagram */ 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 :") << endl; printKDistantfromLeaf(root, 2);}// This code contributed by aashish1995 |
Java
// Java program to print all nodes at a distance k from leaf// A binary tree nodeclass Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }}class BinaryTree { Node root; /* Given a binary tree and a nuber k, print all nodes that are k distant from a leaf*/ int printKDistantfromLeaf(Node node, int k) { if (node == null) return -1; int lk = printKDistantfromLeaf(node.left, k); int rk = printKDistantfromLeaf(node.right, k); boolean isLeaf = lk == -1 && lk == rk; if (lk == 0 || rk == 0 || (isLeaf && k == 0)) System.out.print(" " + node.data); if (isLeaf && k > 0) return k - 1; // leaf node if (lk > 0 && lk < k) return lk - 1; // parent of left leaf if (rk > 0 && rk < k) return rk - 1; // parent of right leaf return -2; } // 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 Vijayan Annamalai |
C#
// C# program to print all nodes at a distance k from leaf// A binary tree nodeusing System;class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }} class BinaryTree { Node root; /* Given a binary tree and a nuber k, print all nodes that are k distant from a leaf*/ int printKDistantfromLeaf(Node node, int k) { if (node == null) return -1; int lk = printKDistantfromLeaf(node.left, k); int rk = printKDistantfromLeaf(node.right, k); bool isLeaf = lk == -1 && lk == rk; if (lk == 0 || rk == 0 || (isLeaf && k == 0)) Console.Write(" " + node.data); if (isLeaf && k > 0) return k - 1; // leaf node if (lk > 0 && lk < k) return lk - 1; // parent of left leaf if (rk > 0 && rk < k) return rk - 1; // parent of right leaf return -2; } // 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.Write("Nodes at distance 2 are :"); tree.printKDistantfromLeaf(tree.root, 2); }}// This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript program to print all nodes// at a distance k from leaf// A binary tree nodeclass Node{ constructor(item) { this.data = item; this.left = null; this.right = null; }}class BinaryTree{ constructor() { this.root = null; } // Given a binary tree and a nuber // k, print all nodes that are k // distant from a leaf printKDistantfromLeaf(node, k) { if (node == null) return -1; var lk = this.printKDistantfromLeaf(node.left, k); var rk = this.printKDistantfromLeaf(node.right, k); var isLeaf = lk == -1 && lk == rk; if (lk == 0 || rk == 0 || (isLeaf && k == 0)) document.write(" " + node.data); // Leaf node if (isLeaf && k > 0) return k - 1; // Parent of left leaf if (lk > 0 && lk < k) return lk - 1; // Parent of right leaf if (rk > 0 && rk < k) return rk - 1; return -2; }}// Driver codevar tree = new BinaryTree();// Let us construct the tree shown// in above diagramtree.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);document.write("Nodes at distance 2 are :<br>");tree.printKDistantfromLeaf(tree.root, 2);// This code is contributed by rdtank</script> |
Output
Nodes at distance 2 are : 3 1
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Another approach :
C++
// Print all nodes that are at distance k from a leaf node#include <iostream>using namespace std;struct bstnode { int data; bstnode* right; bstnode* left;};bstnode* newnode(int data){ bstnode* temp = new bstnode(); temp->data = data; temp->right = temp->left = NULL; return temp;}void del(bstnode* root){ if (root != NULL) { del(root->left); del(root->right); delete root; }}int printk(bstnode* root, int k){ if (root == NULL) return 0; int l = printk(root->left, k); int r = printk(root->right, k); if (l == k || r == k) cout << root->data << " "; return 1 + max(l, r);}int main(){ bstnode* root = NULL; 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); // root->right->right->right=newnode(9); int k = 2; printk(root, k); del(root); return 0;} |
Output
3 1
Time Complexity: O(n) as the code does a simple tree traversal.
Auxiliary Space: O(h) as the function call stack is used where h=height of the tree.
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