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.

Binary Tree

For Example: In the above binary tree, sum of all odd nodes in the path between the nodes $5$ and $6$ will be 5 + 1 + 3 = 9.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 = 5
node2 = 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:

Time Complexity : O(n)
Auxiliary Space : O(n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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