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;
};

// Utility 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;
}

// Utility 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

# Utility 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;
}

// Utility 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

Javascript

<script>

    // JavaScript program to find sum of all odd nodes
    // in the path connecting two given nodes

    // Binary Tree node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    // Utility function to create a
    // new Binary Tree node
    function newNode(data)
    {
          let node = new Node(data);
        return node;
    }

    // Function to check if there is a path from root
    // to the given node. It also populates
    // 'arr' with the given path
    function getPath(root, arr, x)
    {
        // if root is null
        // there is no path
        if (root==null)
            return false;

        // push the node's value in 'arr'
        arr.push(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();
        return false;
    }

    // Function to get the sum of odd nodes in the
    // path between any two nodes in a binary tree
    function sumOddNodes(root, n1, n2)
    {
        // vector to store the path of
        // first node n1 from root
        let path1= [];

        // vector to store the path of
        // second node n2 from root
        let path2= [];

        getPath(root, path1, n1);
        getPath(root, path2, n2);

        let intersection = -1;

        // Get intersection point
        let i = 0, j = 0;
        while (i != path1.length || j != path2.length) {

            // Keep moving forward until no intersection
            // is found
            if (i == j && path1[i] == path2[j]) {
                i++;
                j++;
            }
            else {
                intersection = j - 1;
                break;
            }
        }

        let sum = 0;

        // calculate sum of ODD nodes from the path
        for (i = path1.length - 1; i > intersection; i--)
            if(path1[i]%2!=0)
                sum += path1[i];

        for (i = intersection; i < path2.length; i++)
            if(path2[i]%2!=0)
                sum += path2[i];

        return sum;
    }

    let 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);

    let node1 = 5;
    let node2 = 6;

    document.write(sumOddNodes(root, node1, node2));

</script>

Output

9

Complexity Analysis:

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