Sum of all odd nodes in the path connecting two given nodes
Given a binary tree and two nodes of that binary tree. Find the sum of all nodes with odd values in the path connecting the two given nodes.
For Example: In the above binary tree, sum of all odd nodes in the path between the nodes 
and 
will be 5 + 1 + 3 = 9.
Source : Amazon Interview Experience on Campus
Approach : The idea is to first find the path between the two given nodes of the binary tree using the concept as discussed in: Print path between any two nodes.
Once, we have the path between the two given nodes, calculate sum of all the odd valued nodes in that path and print it.
Below is the implementation of the above approach:
C++
// C++ program to find sum of all odd nodes // in the path connecting two given nodes #include<bits/stdc++.h> using namespace std; // Binary Tree node struct Node { int data; struct Node* left; struct Node* right; }; // Utitlity function to create a // new Binary Tree node struct Node* newNode(int data) { struct Node* node = new Node; node->data = data; node->left = NULL; node->right = NULL; return node; } // Function to check if there is a path from root // to the given node. It also populates // 'arr' with the given path bool getPath(Node* root, vector<int>& arr, int x) { // if root is NULL // there is no path if (!root) return false; // push the node's value in 'arr' arr.push_back(root->data); // if it is the required node // return true if (root->data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (getPath(root->left, arr, x) || getPath(root->right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.pop_back(); return false; } // Function to get the sum of odd nodes in the // path between any two nodes in a binary tree int sumOddNodes(Node* root, int n1, int n2) { // vector to store the path of // first node n1 from root vector<int> path1; // vector to store the path of // second node n2 from root vector<int> path2; getPath(root, path1, n1); getPath(root, path2, n2); int intersection = -1; // Get intersection point int i = 0, j = 0; while (i != path1.size() || j != path2.size()) { // Keep moving forward until no intersection // is found if (i == j && path1[i] == path2[j]) { i++; j++; } else { intersection = j - 1; break; } } int sum = 0; // calculate sum of ODD nodes from the path for (int i = path1.size() - 1; i > intersection; i--) if(path1[i]%2) sum += path1[i]; for (int i = intersection; i < path2.size(); i++) if(path2[i]%2) sum += path2[i]; return sum; } // Driver Code int main() { 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->right = newNode(6); int node1 = 5; int node2 = 6; cout<<sumOddNodes(root, node1, node2); return 0; } |
Java
// Java program to find sum of all odd nodes // in the path connecting two given nodes import java.util.*; class solution { // Binary Tree node static class Node { int data; Node left; Node right; } // Utitlity function to create a // new Binary Tree node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return node; } // Function to check if there is a path from root // to the given node. It also populates // 'arr' with the given path static boolean getPath(Node root, Vector<Integer> arr, int x) { // if root is null // there is no path if (root==null) return false; // push the node's value in 'arr' arr.add(root.data); // if it is the required node // return true if (root.data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (getPath(root.left, arr, x) || getPath(root.right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, remove current node's value from // 'arr'and then return false arr.remove(arr.size()-1); return false; } // Function to get the sum of odd nodes in the // path between any two nodes in a binary tree static int sumOddNodes(Node root, int n1, int n2) { // vector to store the path of // first node n1 from root Vector<Integer> path1= new Vector<Integer>(); // vector to store the path of // second node n2 from root Vector<Integer> path2= new Vector<Integer>(); getPath(root, path1, n1); getPath(root, path2, n2); int intersection = -1; // Get intersection point int i = 0, j = 0; while (i != path1.size() || j != path2.size()) { // Keep moving forward until no intersection // is found if (i == j && path1.get(i) == path2.get(j)) { i++; j++; } else { intersection = j - 1; break; } } int sum = 0; // calculate sum of ODD nodes from the path for (i = path1.size() - 1; i > intersection; i--) if(path1.get(i)%2!=0) sum += path1.get(i); for (i = intersection; i < path2.size(); i++) if(path2.get(i)%2!=0) sum += path2.get(i); return sum; } // Driver Code public static void main(String args[]) { Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(6); int node1 = 5; int node2 = 6; System.out.print(sumOddNodes(root, node1, node2)); } } |
Python3
# Python3 program to find sum of all odd nodes # in the path connecting two given nodes # Binary Tree node class Node: def __init__(self): self.data = 0 self.left = None self.right = None # Utitlity function to create a # Binary Tree node def newNode( data): node = Node() node.data = data node.left = None node.right = None return node # Function to check if there is a path from root # to the given node. It also populates # 'arr' with the given path def getPath(root, arr, x): # if root is None # there is no path if (root == None) : return False # push the node's value in 'arr' arr.append(root.data) # if it is the required node # return True if (root.data == x) : return True # else check whether the required node lies # in the left subtree or right subtree of # the current node if (getPath(root.left, arr, x) or getPath(root.right, arr, x)) : return True # required node does not lie either in the # left or right subtree of the current node # Thus, remove current node's value from # 'arr'and then return False arr.pop() return False # Function to get the sum of odd nodes in the # path between any two nodes in a binary tree def sumOddNodes(root, n1, n2) : # vector to store the path of # first node n1 from root path1 = [] # vector to store the path of # second node n2 from root path2 = [] getPath(root, path1, n1) getPath(root, path2, n2) intersection = -1 # Get intersection point i = 0 j = 0 while (i != len(path1) or j != len(path2)): # Keep moving forward until no intersection # is found if (i == j and path1[i] == path2[j]): i = i + 1 j = j + 1 else : intersection = j - 1 break sum = 0 i = len(path1) - 1 # calculate sum of ODD nodes from the path while ( i > intersection ): if(path1[i] % 2 != 0): sum += path1[i] i = i - 1 i = intersection while ( i < len(path2) ): if(path2[i] % 2 != 0) : sum += path2[i] i = i + 1 return sum # Driver Code root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) root.right.right = newNode(6) node1 = 5node2 = 6 print(sumOddNodes(root, node1, node2)) # This code is contributed by Arnab Kundu |
C#
// C# program to find sum of all odd nodes // in the path connecting two given nodes using System; using System.Collections.Generic; class GFG { // Binary Tree node public class Node { public int data; public Node left; public Node right; } // Utitlity function to create a // new Binary Tree node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return node; } // Function to check if there is a path from // root to the given node. It also populates // 'arr' with the given path static Boolean getPath(Node root, List<int> arr, int x) { // if root is null // there is no path if (root == null) return false; // push the node's value in 'arr' arr.Add(root.data); // if it is the required node // return true if (root.data == x) return true; // else check whether the required node lies // in the left subtree or right subtree of // the current node if (getPath(root.left, arr, x) || getPath(root.right, arr, x)) return true; // required node does not lie either in the // left or right subtree of the current node // Thus, Remove current node's value from // 'arr'and then return false arr.RemoveAt(arr.Count - 1); return false; } // Function to get the sum of odd nodes in the // path between any two nodes in a binary tree static int sumOddNodes(Node root, int n1, int n2) { // List to store the path of // first node n1 from root List<int> path1 = new List<int>(); // List to store the path of // second node n2 from root List<int> path2 = new List<int>(); getPath(root, path1, n1); getPath(root, path2, n2); int intersection = -1; // Get intersection point int i = 0, j = 0; while (i < path1.Count || j < path2.Count) { // Keep moving forward until // no intersection is found if ( i == j && path1[i] == path2[j]) { i++; j++; } else { intersection = j - 1; break; } } int sum = 0; // calculate sum of ODD nodes from the path for (i = path1.Count - 1; i > intersection; i--) if(path1[i] % 2 != 0) sum += path1[i]; for (i = intersection; i < path2.Count; i++) if(path2[i] % 2 != 0) sum += path2[i]; return sum; } // Driver Code public static void Main(String []args) { Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(6); int node1 = 5; int node2 = 6; Console.Write(sumOddNodes(root, node1, node2)); } } // This code is contributed by Arnub Kundu |
Output:
9
Time Complexity : O(n)
Auxiliary Space : O(n)
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