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

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

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

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

Method 1
We can solve this problem by considering the fact that both node and its children can’t be in sum at same time, so when we take a node into our sum we will call recursively for its grandchildren or when we don’t take this node we will call for all its children nodes and finally we will choose maximum from both of these results.
It can be seen easily that above approach can lead to solving same subproblem many times, for example in 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 below code a map is used for memoizing the result which stores result of 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 map is returned directly.
Please see below code for better understanding.

## C++

 `// C++ program to find maximum sum from a subset of ` `// nodes of binary tree ` `#include ` `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 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 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 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 `

## 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 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 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 mp=``new` `Dictionary(); ` `        ``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 `

Output:

```11
```

Method 2 (Using pair)
Return a pair for each node in the binary tree such that first of the pair indicates maximum sum when the data of node is included and 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 ` `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 */`

Output:

```21
```

Time complexity: O(n)

Thanks to Surbhi Rastogi for suggesting this method.

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