Given a perfect binary tree, print nodes of middle level without computing its height. A perfect binary tree is a binary tree in which all interior nodes have two children and all leaves have the same depth or same level.
Output : 4 5 6 7
The idea is similar to method 2 of finding middle of singly linked list.
Use fast and slow (or tortoise) pointers in each route of the tree.
- Advance fast pointer towards leaf by 2.
- Advance slow pointer towards lead by 1.
- If fast pointer reaches the leaf print value at the slow pointer
- Check if the fast->left->left exists, then recursively move slow pointer by one step and fast pointer by two steps.
- If the fast->left->left doesn’t exist (in case of even number of levels), the move both the pointers by one step.
C++
#include <bits/stdc++.h> using namespace std; /* A binary tree node has key, pointer to left child and a pointer to right child */struct Node { int key; struct Node *left, *right; }; /* To create a newNode of tree and return pointer */struct Node* newNode(int key) { Node* temp = new Node; temp->key = key; temp->left = temp->right = NULL; return (temp); } // Takes two parameters - same initially and // calls recursively void printMiddleLevelUtil(Node* a, Node* b) { // Base case e if (a == NULL || b == NULL) return; // Fast pointer has reached the leaf so print // value at slow pointer if ((b->left == NULL) && (b->right == NULL)) { cout << a->key << " "; return; } // Recursive call // root.left.left and root.left.right will // print same value // root.right.left and root.right.right // will print same value // So we use any one of the condition if (b->left->left) { printMiddleLevelUtil(a->left, b->left->left); printMiddleLevelUtil(a->right, b->left->left); } else { printMiddleLevelUtil(a->left, b->left); printMiddleLevelUtil(a->right, b->left); } } // Main printing method that take a Tree as input void printMiddleLevel(Node* node) { printMiddleLevelUtil(node, node); } // Driver program to test above functions int main() { Node* n1 = newNode(1); Node* n2 = newNode(2); Node* n3 = newNode(3); Node* n4 = newNode(4); Node* n5 = newNode(5); Node* n6 = newNode(6); Node* n7 = newNode(7); n2->left = n4; n2->right = n5; n3->left = n6; n3->right = n7; n1->left = n2; n1->right = n3; printMiddleLevel(n1); } // This code is contributed by Prasad Kshirsagar |
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Java
// Tree node definition class Node { public int key; public Node left; public Node right; public Node(int val) { this.left = null; this.right = null; this.key = val; } } public class PrintMiddle { // Takes two parameters - same initially and // calls recursively private static void printMiddleLevelUtil(Node a, Node b) { // Base case e if (a == null || b == null) return; // Fast pointer has reached the leaf so print // value at slow pointer if ((b.left == null) && (b.right == null)) { System.out.print(a.key + " "); return; } // Recursive call // root.left.left and root.left.right will // print same value // root.right.left and root.right.right // will print same value // So we use any one of the condition if (b.left.left) { printMiddleLevelUtil(a.left, b.left.left); printMiddleLevelUtil(a.right, b.left.left); } else { printMiddleLevelUtil(a.left, b.left); printMiddleLevelUtil(a.right, b.left); } } // Main printing method that take a Tree as input public static void printMiddleLevel(Node node) { printMiddleLevelUtil(node, node); } // Driver code public static void main(String[] args) { Node n1 = new Node(1); Node n2 = new Node(2); Node n3 = new Node(3); Node n4 = new Node(4); Node n5 = new Node(5); Node n6 = new Node(6); Node n7 = new Node(7); n2.left = n4; n2.right = n5; n3.left = n6; n3.right = n7; n1.left = n2; n1.right = n3; printMiddleLevel(n1); } } |
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C#
using System; // Tree node definition public class Node { public int key; public Node left; public Node right; public Node(int val) { this.left = null; this.right = null; this.key = val; } } public class PrintMiddle { // Takes two parameters - same initially and // calls recursively private static void printMiddleLevelUtil(Node a, Node b) { // Base case e if (a == null || b == null) return; // Fast pointer has reached the leaf so print // value at slow pointer if ((b.left == null) && (b.right == null)) { Console.Write(a.key + " "); return; } // Recursive call // root.left.left and root.left.right will // print same value // root.right.left and root.right.right // will print same value // So we use any one of the condition if (b.left.left!=null) { printMiddleLevelUtil(a.left, b.left.left); printMiddleLevelUtil(a.right, b.left.left); } else { printMiddleLevelUtil(a.left, b.left); printMiddleLevelUtil(a.right, b.left); } } // Main printing method that take a Tree as input public static void printMiddleLevel(Node node) { printMiddleLevelUtil(node, node); } // Driver code public static void Main(String[] args) { Node n1 = new Node(1); Node n2 = new Node(2); Node n3 = new Node(3); Node n4 = new Node(4); Node n5 = new Node(5); Node n6 = new Node(6); Node n7 = new Node(7); n2.left = n4; n2.right = n5; n3.left = n6; n3.right = n7; n1.left = n2; n1.right = n3; printMiddleLevel(n1); } } // This code is contributed by Amit Katiyar |
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Output
2 3
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