A tree is Continuous tree if in each root to leaf path, absolute difference between keys of two adjacent is 1. We are given a binary tree, we need to check if tree is continuous or not.
Examples:
Input : 3
/
2 4
/
1 3 5
Output: "Yes"
// 3->2->1 every two adjacent node's absolute difference is 1
// 3->2->3 every two adjacent node's absolute difference is 1
// 3->4->5 every two adjacent node's absolute difference is 1
Input : 7
/
5 8
/
6 4 10
Output: "No"
// 7->5->6 here absolute difference of 7 and 5 is not 1.
// 7->5->4 here absolute difference of 7 and 5 is not 1.
// 7->8->10 here absolute difference of 8 and 10 is not 1.
The solution requires a traversal of tree. The important things to check are to make sure that all corner cases are handled. The corner cases include, empty tree, single node tree, a node with only left child and a node with only right child.
In tree traversal, we recursively check if left and right subtree are continuous. We also check if difference between keys of current node’s key and its children keys is 1. Below is the implementation of the idea.
C++
// C++ program to check if a tree is continuous or not #include<bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; struct Node* left, * right; }; /* 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 = new Node; node->data = data; node->left = node->right = NULL; return(node); } // Function to check tree is continuous or not bool treeContinuous(struct Node *ptr) { // if next node is empty then return true if (ptr == NULL) return true; // if current node is leaf node then return true // because it is end of root to leaf path if (ptr->left == NULL && ptr->right == NULL) return true; // If left subtree is empty, then only check right if (ptr->left == NULL) return (abs(ptr->data - ptr->right->data) == 1) && treeContinuous(ptr->right); // If right subtree is empty, then only check left if (ptr->right == NULL) return (abs(ptr->data - ptr->left->data) == 1) && treeContinuous(ptr->left); // If both left and right subtrees are not empty, check // everything return abs(ptr->data - ptr->left->data)==1 && abs(ptr->data - ptr->right->data)==1 && treeContinuous(ptr->left) && treeContinuous(ptr->right); } /* Driver program to test mirror() */int main() { struct Node *root = newNode(3); root->left = newNode(2); root->right = newNode(4); root->left->left = newNode(1); root->left->right = newNode(3); root->right->right = newNode(5); treeContinuous(root)? cout << "Yes" : cout << "No"; return 0; } |
Java
// Java program to check if a tree is continuous or not import java.util.*; class solution { /* A binary tree node has data, pointer to left child and a pointer to right child */static class Node { int data; Node left, right; }; /* Helper function that allocates a new node with the given data and null left and right pointers. */ static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return(node); } // Function to check tree is continuous or not static boolean treeContinuous( Node ptr) { // if next node is empty then return true if (ptr == null) return true; // if current node is leaf node then return true // because it is end of root to leaf path if (ptr.left == null && ptr.right == null) return true; // If left subtree is empty, then only check right if (ptr.left == null) return (Math.abs(ptr.data - ptr.right.data) == 1) && treeContinuous(ptr.right); // If right subtree is empty, then only check left if (ptr.right == null) return (Math.abs(ptr.data - ptr.left.data) == 1) && treeContinuous(ptr.left); // If both left and right subtrees are not empty, check // everything return Math.abs(ptr.data - ptr.left.data)==1 && Math.abs(ptr.data - ptr.right.data)==1 && treeContinuous(ptr.left) && treeContinuous(ptr.right); } /* Driver program to test mirror() */public static void main(String args[]) { Node root = newNode(3); root.left = newNode(2); root.right = newNode(4); root.left.left = newNode(1); root.left.right = newNode(3); root.right.right = newNode(5); if(treeContinuous(root)) System.out.println( "Yes") ; else System.out.println( "No"); } } //contributed by Arnab Kundu |
C#
// C# program to check if a tree is continuous or not using System; class solution { /* A binary tree node has data, pointer to left child and a pointer to right child */class Node { public int data; public Node left, right; }; /* Helper function that allocates a new node with the given data and null left and right pointers. */ static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return(node); } // Function to check tree is continuous or not static Boolean treeContinuous( Node ptr) { // if next node is empty then return true if (ptr == null) return true; // if current node is leaf node then return true // because it is end of root to leaf path if (ptr.left == null && ptr.right == null) return true; // If left subtree is empty, then only check right if (ptr.left == null) return (Math.Abs(ptr.data - ptr.right.data) == 1) && treeContinuous(ptr.right); // If right subtree is empty, then only check left if (ptr.right == null) return (Math.Abs(ptr.data - ptr.left.data) == 1) && treeContinuous(ptr.left); // If both left and right subtrees are not empty, check // everything return Math.Abs(ptr.data - ptr.left.data)==1 && Math.Abs(ptr.data - ptr.right.data)==1 && treeContinuous(ptr.left) && treeContinuous(ptr.right); } /* Driver program to test mirror() */static public void Main(String []args) { Node root = newNode(3); root.left = newNode(2); root.right = newNode(4); root.left.left = newNode(1); root.left.right = newNode(3); root.right.right = newNode(5); if(treeContinuous(root)) Console.WriteLine( "Yes") ; else Console.WriteLine( "No"); } } //contributed by Arnab Kundu |
Output:
Yes
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