Maximum sum of nodes in Binary tree such that no two are adjacent

Difficulty Level : Hard
Last Updated : 10 Aug, 2021

Given a binary tree with a value associated with each node, we need to choose a subset of these nodes such that the sum of chosen nodes is maximum under a constraint that no two chosen nodes in the subset should be directly connected that is, if we have taken a node in our sum then we can’t take any of its children in consideration and vice versa.

Examples:

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In above binary tree chosen nodes are encircled 
and are not directly connected and their sum is
maximum possible.

Method 1
We can solve this problem by considering the fact that both node and its children can’t be in sum at the same time, so when we take a node into our sum we will call recursively for its grandchildren or if we don’t take this node then we will call for all its children nodes and finally we will choose maximum from both of the results.
It can be seen easily that the above approach can lead to solving the same subproblem many times, for example in the above diagram node 1 calls node 4 and 5 when its value is chosen and node 3 also calls them when its value is not chosen so these nodes are processed more than once. We can stop solving these nodes more than once by memoizing the result at all nodes.
In the below code, a map is used for memorizing the result which stores the result of the complete subtree rooted at a node in the map so that if it is called again, the value is not calculated again instead stored value from the map is returned directly.

Please see the below code for a better understanding.

C++

// C++ program to find maximum sum from a subset of
// nodes of binary tree
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node structure */
struct node
{
    int data;
    struct node *left, *right;
};
 
/* Utility function to create a new Binary Tree node */
struct node* newNode(int data)
{
    struct node *temp = new struct node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
//  Declaration of methods
int sumOfGrandChildren(node* node);
int getMaxSum(node* node);
int getMaxSumUtil(node* node, map<struct node*, int>& mp);
 
// method returns maximum sum possible from subtrees rooted
// at grandChildrens of node 'node'
int sumOfGrandChildren(node* node, map<struct node*, int>& mp)
{
    int sum = 0;
 
    //  call for children of left child only if it is not NULL
    if (node->left)
        sum += getMaxSumUtil(node->left->left, mp) +
               getMaxSumUtil(node->left->right, mp);
 
    //  call for children of right child only if it is not NULL
    if (node->right)
        sum += getMaxSumUtil(node->right->left, mp) +
               getMaxSumUtil(node->right->right, mp);
 
    return sum;
}
 
//  Utility method to return maximum sum rooted at node 'node'
int getMaxSumUtil(node* node, map<struct node*, int>& mp)
{
    if (node == NULL)
        return 0;
 
    // If node is already processed then return calculated
    // value from map
    if (mp.find(node) != mp.end())
        return mp[node];
 
    //  take current node value and call for all grand children
    int incl = node->data + sumOfGrandChildren(node, mp);
 
    //  don't take current node value and call for all children
    int excl = getMaxSumUtil(node->left, mp) +
               getMaxSumUtil(node->right, mp);
 
    //  choose maximum from both above calls and store that in map
    mp[node] = max(incl, excl);
 
    return mp[node];
}
 
// Returns maximum sum from subset of nodes
// of binary tree under given constraints
int getMaxSum(node* node)
{
    if (node == NULL)
        return 0;
    map<struct node*, int> mp;
    return getMaxSumUtil(node, mp);
}
 
//  Driver code to test above methods
int main()
{
    node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->right->left = newNode(4);
    root->right->right = newNode(5);
    root->left->left = newNode(1);
 
    cout << getMaxSum(root) << endl;
    return 0;
}

Java

// Java program to find maximum sum from a subset of
// nodes of binary tree
import java.util.HashMap;
public class FindSumOfNotAdjacentNodes {
 
    // method returns maximum sum possible from subtrees rooted
    // at grandChildrens of node 'node'
    public static int sumOfGrandChildren(Node node, HashMap<Node,Integer> mp)
    {
        int sum = 0;
        //  call for children of left child only if it is not NULL
        if (node.left!=null)
            sum += getMaxSumUtil(node.left.left, mp) +
                   getMaxSumUtil(node.left.right, mp);
   
        //  call for children of right child only if it is not NULL
        if (node.right!=null)
            sum += getMaxSumUtil(node.right.left, mp) +
                   getMaxSumUtil(node.right.right, mp);
        return sum;
    }
 
    //  Utility method to return maximum sum rooted at node 'node'
    public static int getMaxSumUtil(Node node, HashMap<Node,Integer> mp)
    {
        if (node == null)
            return 0;
   
        // If node is already processed then return calculated
        // value from map
        if(mp.containsKey(node))
            return mp.get(node);
   
        //  take current node value and call for all grand children
        int incl = node.data + sumOfGrandChildren(node, mp);
   
        //  don't take current node value and call for all children
        int excl = getMaxSumUtil(node.left, mp) +
                   getMaxSumUtil(node.right, mp);
   
        //  choose maximum from both above calls and store that in map
        mp.put(node,Math.max(incl, excl));
   
        return mp.get(node);
    }
 
    // Returns maximum sum from subset of nodes
    // of binary tree under given constraints
    public static int getMaxSum(Node node)
    {
        if (node == null)
            return 0;
        HashMap<Node,Integer> mp=new HashMap<>();
        return getMaxSumUtil(node, mp);
    }
 
    public static void main(String args[])
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.right.left = new Node(4);
        root.right.right = new Node(5);
        root.left.left = new Node(1);    
        System.out.print(getMaxSum(root));
    }
}
 
/* A binary tree node structure */
class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data=data;
        left=right=null;
    }
};
//This code is contributed by Gaurav Tiwari

Python3

# Python3 program to find
# maximum sum from a subset
# of nodes of binary tree
 
# A binary tree node structure
class Node:
     
    def __init__(self, data):
     
        self.data = data
        self.left = None
        self.right = None
 
# Utility function to create
# a new Binary Tree node
def newNode(data):
 
    temp = Node(data)
    return temp;
 
# method returns maximum sum
# possible from subtrees rooted
# at grandChildrens of node 'node'
def sumOfGrandChildren(node, mp):
 
    sum = 0;
 
    # call for children of left
    # child only if it is not NULL
    if (node.left):
        sum += (getMaxSumUtil(node.left.left, mp) +
                getMaxSumUtil(node.left.right, mp));
 
    # call for children of right
    # child only if it is not NULL
    if (node.right):
        sum += (getMaxSumUtil(node.right.left, mp) +
                getMaxSumUtil(node.right.right, mp));
 
    return sum;
 
# Utility method to return
# maximum sum rooted at node
# 'node'
def getMaxSumUtil(node, mp):
 
    if (node == None):
        return 0;
 
    # If node is already processed
    # then return calculated
    # value from map
    if node in mp:
        return mp[node];
 
    # take current node value
    # and call for all grand children
    incl = (node.data +
            sumOfGrandChildren(node, mp));
 
    # don't take current node
    # value and call for all children
    excl = (getMaxSumUtil(node.left, mp) +
            getMaxSumUtil(node.right, mp));
 
    # choose maximum from both
    # above calls and store that
    # in map
    mp[node] = max(incl, excl);
 
    return mp[node];
 
# Returns maximum sum from
# subset of nodes of binary
# tree under given constraints
def getMaxSum(node):
 
    if (node == None):
        return 0;
     
    mp = dict()
    return getMaxSumUtil(node, mp);
 
# Driver code
if __name__=="__main__":
     
    root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.right.left = newNode(4);
    root.right.right = newNode(5);
    root.left.left = newNode(1);
     
    print(getMaxSum(root))
     
# This code is contributed by Rutvik_56

C#

// C# program to find maximum sum from a subset of
// nodes of binary tree
using System;
using System.Collections.Generic;
 
public class FindSumOfNotAdjacentNodes
{
 
    // method returns maximum sum
    // possible from subtrees rooted
    // at grandChildrens of node 'node'
    public static int sumOfGrandChildren(Node node,
                            Dictionary<Node,int> mp)
    {
        int sum = 0;
         
        // call for children of left
        // child only if it is not NULL
        if (node.left != null)
            sum += getMaxSumUtil(node.left.left, mp) +
                getMaxSumUtil(node.left.right, mp);
     
        // call for children of right
        // child only if it is not NULL
        if (node.right != null)
            sum += getMaxSumUtil(node.right.left, mp) +
                getMaxSumUtil(node.right.right, mp);
        return sum;
    }
 
    // Utility method to return maximum
    // sum rooted at node 'node'
    public static int getMaxSumUtil(Node node,
                        Dictionary<Node,int> mp)
    {
        if (node == null)
            return 0;
     
        // If node is already processed then
        // return calculated value from map
        if(mp.ContainsKey(node))
            return mp[node];
     
        // take current node value and
        // call for all grand children
        int incl = node.data + sumOfGrandChildren(node, mp);
     
        // don't take current node value and
        // call for all children
        int excl = getMaxSumUtil(node.left, mp) +
                getMaxSumUtil(node.right, mp);
     
        // choose maximum from both above
        // calls and store that in map
        mp.Add(node,Math.Max(incl, excl));
     
        return mp[node];
    }
 
    // Returns maximum sum from subset of nodes
    // of binary tree under given constraints
    public static int getMaxSum(Node node)
    {
        if (node == null)
            return 0;
        Dictionary<Node,int> mp=new Dictionary<Node,int>();
        return getMaxSumUtil(node, mp);
    }
 
    // Driver code
    public static void Main(String []args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.right.left = new Node(4);
        root.right.right = new Node(5);
        root.left.left = new Node(1);    
        Console.Write(getMaxSum(root));
    }
}
 
/* A binary tree node structure */
public class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data=data;
        left=right=null;
    }
};
 
// This code has been contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program to find maximum
// sum from a subset of nodes of binary tree
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}
 
// Method returns maximum sum possible
// from subtrees rooted at grandChildrens
// of node 'node'
function sumOfGrandChildren(node, mp)
{
    let sum = 0;
     
    // Call for children of left child
    // only if it is not NULL
    if (node.left!=null)
        sum += getMaxSumUtil(node.left.left, mp) +
               getMaxSumUtil(node.left.right, mp);
 
    // Call for children of right child
    // only if it is not NULL
    if (node.right!=null)
        sum += getMaxSumUtil(node.right.left, mp) +
               getMaxSumUtil(node.right.right, mp);
    return sum;
}
 
// Utility method to return maximum
// sum rooted at node 'node'
function getMaxSumUtil(node, mp)
{
    if (node == null)
        return 0;
 
    // If node is already processed then return
    // calculated value from map
    if (mp.has(node))
        return mp.get(node);
 
    // Take current node value and call for
    // all grand children
    let incl = node.data + sumOfGrandChildren(node, mp);
 
    // Don't take current node value and call
    // for all children
    let excl = getMaxSumUtil(node.left, mp) +
               getMaxSumUtil(node.right, mp);
 
    // Choose maximum from both above
    // calls and store that in map
    mp.set(node,Math.max(incl, excl));
 
    return mp.get(node);
}
 
// Returns maximum sum from subset of nodes
// of binary tree under given constraints
function getMaxSum(node)
{
    if (node == null)
        return 0;
         
    let mp = new Map();
    return getMaxSumUtil(node, mp);
}
 
// Driver code
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.right.left = new Node(4);
root.right.right = new Node(5);
root.left.left = new Node(1);   
 
document.write(getMaxSum(root));
 
// This code is contributed by divyeshrabadiya07
 
</script>

Output:

Method 2 (Using pair)
Return a pair for each node in the binary tree such that the first of the pair indicates maximum sum when the data of a node is included and the second indicates maximum sum when the data of a particular node is not included.

C++

// C++ program to find maximum sum in Binary Tree
// such that no two nodes are adjacent.
#include<iostream>
using namespace std;
 
class Node
{
public:
    int data;
    Node* left, *right;
    Node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};
 
pair<int, int> maxSumHelper(Node *root)
{
    if (root==NULL)
    {
        pair<int, int> sum(0, 0);
        return sum;
    }
    pair<int, int> sum1 = maxSumHelper(root->left);
    pair<int, int> sum2 = maxSumHelper(root->right);
    pair<int, int> sum;
 
    // This node is included (Left and right children
    // are not included)
    sum.first = sum1.second + sum2.second + root->data;
 
    // This node is excluded (Either left or right
    // child is included)
    sum.second = max(sum1.first, sum1.second) +
                 max(sum2.first, sum2.second);
 
    return sum;
}
 
int maxSum(Node *root)
{
    pair<int, int> res = maxSumHelper(root);
    return max(res.first, res.second);
}
 
// Driver code
int main()
{
    Node *root= new Node(10);
    root->left= new Node(1);
    root->left->left= new Node(2);
    root->left->left->left= new Node(1);
    root->left->right= new Node(3);
    root->left->right->left= new Node(4);
    root->left->right->right= new Node(5);
    cout << maxSum(root);
    return 0;
}

Java

// Java program to find maximum sum in Binary Tree
// such that no two nodes are adjacent.
public class FindSumOfNotAdjacentNodes {
 
    public static Pair maxSumHelper(Node root)
    {
        if (root==null)
        {
            Pair sum=new Pair(0, 0);
            return sum;
        }
        Pair sum1 = maxSumHelper(root.left);
        Pair sum2 = maxSumHelper(root.right);
        Pair sum=new Pair(0,0);
   
        // This node is included (Left and right children
        // are not included)
        sum.first = sum1.second + sum2.second + root.data;
   
        // This node is excluded (Either left or right
        // child is included)
        sum.second = Math.max(sum1.first, sum1.second) +
                     Math.max(sum2.first, sum2.second);
   
        return sum;
    }
 
    // Returns maximum sum from subset of nodes
    // of binary tree under given constraints
    public static int maxSum(Node root)
    {
        Pair res=maxSumHelper(root);
        return Math.max(res.first, res.second);
    }
 
    public static void main(String args[]) {
        Node root= new Node(10);
        root.left= new Node(1);
        root.left.left= new Node(2);
        root.left.left.left= new Node(1);
        root.left.right= new Node(3);
        root.left.right.left= new Node(4);
        root.left.right.right= new Node(5);
        System.out.print(maxSum(root));
    }
}
 
/* A binary tree node structure */
class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data=data;
        left=right=null;
    }
};
 
/* Pair class */
class Pair
{
    int first,second;
    Pair(int first,int second)
    {
        this.first=first;
        this.second=second;
    }
}
//This code is contributed by Gaurav Tiwari

Python3

# Python3 program to find maximum sum in Binary
# Tree such that no two nodes are adjacent.
 
# Binary Tree Node
 
""" utility that allocates a newNode
with the given key """
class newNode:
 
    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
def maxSumHelper(root) :
 
    if (root == None):
     
        sum = [0, 0]
        return sum
     
    sum1 = maxSumHelper(root.left)
    sum2 = maxSumHelper(root.right)
    sum = [0, 0]
 
    # This node is included (Left and right
    # children are not included)
    sum[0] = sum1[1] + sum2[1] + root.data
 
    # This node is excluded (Either left or
    # right child is included)
    sum[1] = (max(sum1[0], sum1[1]) +
              max(sum2[0], sum2[1]))
 
    return sum
 
def maxSum(root) :
 
    res = maxSumHelper(root)
    return max(res[0], res[1])
 
# Driver Code
if __name__ == '__main__':
    root = newNode(10)
    root.left = newNode(1)
    root.left.left = newNode(2)
    root.left.left.left = newNode(1)
    root.left.right = newNode(3)
    root.left.right.left = newNode(4)
    root.left.right.right = newNode(5)
    print(maxSum(root))
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to find maximum sum in Binary Tree
// such that no two nodes are adjacent.
using System;
 
public class FindSumOfNotAdjacentNodes
{
 
    public static Pair maxSumHelper(Node root)
    {
        Pair sum;
        if (root == null)
        {
            sum=new Pair(0, 0);
            return sum;
        }
        Pair sum1 = maxSumHelper(root.left);
        Pair sum2 = maxSumHelper(root.right);
        Pair sum3 = new Pair(0,0);
     
        // This node is included (Left and
        // right children are not included)
        sum3.first = sum1.second + sum2.second + root.data;
     
        // This node is excluded (Either left
        // or right child is included)
        sum3.second = Math.Max(sum1.first, sum1.second) +
                    Math.Max(sum2.first, sum2.second);
     
        return sum3;
    }
 
    // Returns maximum sum from subset of nodes
    // of binary tree under given constraints
    public static int maxSum(Node root)
    {
        Pair res=maxSumHelper(root);
        return Math.Max(res.first, res.second);
    }
 
    // Driver code
    public static void Main()
    {
        Node root = new Node(10);
        root.left = new Node(1);
        root.left.left = new Node(2);
        root.left.left.left = new Node(1);
        root.left.right = new Node(3);
        root.left.right.left = new Node(4);
        root.left.right.right = new Node(5);
        Console.Write(maxSum(root));
    }
}
 
/* A binary tree node structure */
public class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
};
 
/* Pair class */
public class Pair
{
    public int first,second;
    public Pair(int first,int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
/* This code is contributed PrinciRaj1992 */

Javascript

<script>
 
// JavaScript program to find maximum sum in Binary Tree
// such that no two nodes are adjacent.
 
/* A binary tree node structure */
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
/* Pair class */
class Pair
{
    constructor(first, second)
    {
        this.first = first;
        this.second = second;
    }
}
 
function maxSumHelper(root)
{
    var sum;
    if (root == null)
    {
        sum=new Pair(0, 0);
        return sum;
    }
    var sum1 = maxSumHelper(root.left);
    var sum2 = maxSumHelper(root.right);
    var sum3 = new Pair(0,0);
 
    // This node is included (Left and
    // right children are not included)
    sum3.first = sum1.second + sum2.second + root.data;
 
    // This node is excluded (Either left
    // or right child is included)
    sum3.second = Math.max(sum1.first, sum1.second) +
                Math.max(sum2.first, sum2.second);
 
    return sum3;
}
// Returns maximum sum from subset of nodes
// of binary tree under given constraints
function maxSum(root)
{
    var res=maxSumHelper(root);
    return Math.max(res.first, res.second);
}
// Driver code
var root = new Node(10);
root.left = new Node(1);
root.left.left = new Node(2);
root.left.left.left = new Node(1);
root.left.right = new Node(3);
root.left.right.left = new Node(4);
root.left.right.right = new Node(5);
document.write(maxSum(root));
 
 
</script>

Output:

Time complexity: O(n)
Thanks to Surbhi Rastogi for suggesting this method.

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