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# Sum of nodes in top view of binary tree

Top view of a binary tree is the set of nodes visible when the tree is viewed from the top. Given a binary tree, the task is to print the sum of nodes in top view.
Examples:

```Input:
1
/  \
2    3
/ \    \
4   5    6

Output: 16

Input:
1
/  \
2      3
\
4
\
5
\
6

Output: 12```

Approach: The idea is to put nodes of same horizontal distance together. We do a level order traversal so that the topmost node at a horizontal node is visited before any other node of same horizontal distance below it and we keep summing up their values and store the result in variable sum. Hashing is used to check if a node at given horizontal distance is seen or not.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Structure of binary tree``struct` `Node {``    ``Node* left;``    ``Node* right;``    ``int` `hd;``    ``int` `data;``};` `// Function to create a new node``Node* newNode(``int` `key)``{``    ``Node* node = ``new` `Node();``    ``node->left = node->right = NULL;``    ``node->data = key;``    ``return` `node;``}` `// Function that returns the sum of``// nodes in top view of binary tree``int` `SumOfTopView(Node* root)``{``    ``if` `(root == NULL)``        ``return` `0;` `    ``queue q;` `    ``map<``int``, ``int``> m;``    ``int` `hd = 0;` `    ``root->hd = hd;` `    ``int` `sum = 0;` `    ``// Push node and horizontal distance to queue``    ``q.push(root);` `    ``while` `(q.size()) {``        ``hd = root->hd;` `        ``// Count function returns 1 if the container``        ``// contains an element whose key is equivalent``        ``// to hd, or returns zero otherwise.``        ``if` `(m.count(hd) == 0) {``            ``m[hd] = root->data;``            ``sum += m[hd];``        ``}``        ``if` `(root->left) {``            ``root->left->hd = hd - 1;``            ``q.push(root->left);``        ``}``        ``if` `(root->right) {``            ``root->right->hd = hd + 1;``            ``q.push(root->right);``        ``}``        ``q.pop();``        ``root = q.front();``    ``}` `    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->right = newNode(4);``    ``root->left->right->right = newNode(5);``    ``root->left->right->right->right = newNode(6);` `    ``cout << SumOfTopView(root);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;``class` `Sol``{``    ` `// Structure of binary tree``static` `class` `Node``{``    ``Node left;``    ``Node right;``    ``int` `hd;``    ``int` `data;``};` `// Function to create a new node``static` `Node newNode(``int` `key)``{``    ``Node node = ``new` `Node();``    ``node.left = node.right = ``null``;``    ``node.data = key;``    ``return` `node;``}` `// Function that returns the sum of``// nodes in top view of binary tree``static` `int` `SumOfTopView(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``Queue q = ``new` `LinkedList();` `    ``Map m = ``new` `HashMap();``    ``int` `hd = ``0``;` `    ``root.hd = hd;` `    ``int` `sum = ``0``;` `    ``// Push node and horizontal distance to queue``    ``q.add(root);` `    ``while` `(q.size() > ``0``)``    ``{``        ``hd = root.hd;` `        ``// Count function returns 1 if the container``        ``// contains an element whose key is equivalent``        ``// to hd, or returns zero otherwise.``        ``if` `(!m.containsKey(hd))``        ``{``            ``m.put(hd, root.data);``            ``sum += m.get(hd);``        ``}``        ``if` `(root.left != ``null``)``        ``{``            ``root.left.hd = hd - ``1``;``            ``q.add(root.left);``        ``}``        ``if` `(root.right != ``null``)``        ``{``            ``root.right.hd = hd + ``1``;``            ``q.add(root.right);``        ``}``        ``q.remove();``        ``if``(q.size() > ``0``)``            ``root = q.peek();``    ``}` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``Node root = newNode(``1``);``    ``root.left = newNode(``2``);``    ``root.right = newNode(``3``);``    ``root.left.right = newNode(``4``);``    ``root.left.right.right = newNode(``5``);``    ``root.left.right.right.right = newNode(``6``);` `    ``System.out.print(SumOfTopView(root));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach``from` `collections ``import` `defaultdict` `class` `Node:``    ` `    ``def` `__init__(``self``, key):``        ``self``.data ``=` `key``        ``self``.hd ``=` `None``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function that returns the sum of``# nodes in top view of binary tree``def` `SumOfTopView(root):` `    ``if` `root ``=``=` `None``:``        ``return` `0` `    ``q ``=` `[]` `    ``m ``=` `defaultdict(``lambda``:``0``)``    ``hd, ``Sum` `=` `0``, ``0` `    ``root.hd ``=` `hd` `    ``# Push node and horizontal``    ``# distance to queue``    ``q.append(root)` `    ``while` `len``(q) > ``0``:``        ``hd ``=` `root.hd` `        ``# Count function returns 1 if``        ``# the container contains an``        ``# element whose key is equivalent``        ``# to hd, or returns zero otherwise.``        ``if` `m[hd] ``=``=` `0``:``            ``m[hd] ``=` `root.data``            ``Sum` `+``=` `m[hd]``        ` `        ``if` `root.left !``=` `None``:``            ``root.left.hd ``=` `hd ``-` `1``            ``q.append(root.left)``        ` `        ``if` `root.right !``=` `None``:``            ``root.right.hd ``=` `hd ``+` `1``            ``q.append(root.right)``        ` `        ``q.pop(``0``)``        ``if` `len``(q) > ``0``:``            ``root ``=` `q[``0``]``    ` `    ``return` `Sum` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``root ``=` `Node(``1``)``    ``root.left ``=` `Node(``2``)``    ``root.right ``=` `Node(``3``)``    ``root.left.right ``=` `Node(``4``)``    ``root.left.right.right ``=` `Node(``5``)``    ``root.left.right.right.right ``=` `Node(``6``)` `    ``print``(SumOfTopView(root))` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `// Structure of binary tree``public` `class` `Node``{``    ``public` `Node left;``    ``public` `Node right;``    ``public` `int` `hd;``    ``public` `int` `data;``};` `// Function to create a new node``static` `Node newNode(``int` `key)``{``    ``Node node = ``new` `Node();``    ``node.left = node.right = ``null``;``    ``node.data = key;``    ``return` `node;``}` `// Function that returns the sum of``// nodes in top view of binary tree``static` `int` `SumOfTopView(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0;` `    ``Queue q = ``new` `Queue();` `    ``Dictionary<``int``,``               ``int``> m = ``new` `Dictionary<``int``,``                                       ``int``>();``    ``int` `hd = 0;` `    ``root.hd = hd;` `    ``int` `sum = 0;` `    ``// Push node and horizontal distance to queue``    ``q.Enqueue(root);` `    ``while` `(q.Count > 0)``    ``{``        ``hd = root.hd;` `        ``// Count function returns 1 if the container``        ``// contains an element whose key is equivalent``        ``// to hd, or returns zero otherwise.``        ``if` `(!m.ContainsKey(hd))``        ``{``            ``m.Add(hd, root.data);``            ``sum += m[hd];``        ``}``        ` `        ``if` `(root.left != ``null``)``        ``{``            ``root.left.hd = hd - 1;``            ``q.Enqueue(root.left);``        ``}``        ` `        ``if` `(root.right != ``null``)``        ``{``            ``root.right.hd = hd + 1;``            ``q.Enqueue(root.right);``        ``}``        ``q.Dequeue();``        ``if``(q.Count > 0)``            ``root = q.Peek();``    ``}``    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``Node root = newNode(1);``    ``root.left = newNode(2);``    ``root.right = newNode(3);``    ``root.left.right = newNode(4);``    ``root.left.right.right = newNode(5);``    ``root.left.right.right.right = newNode(6);` `    ``Console.Write(SumOfTopView(root));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`12`

Time Complexity: O(N)

Auxiliary Space: O(N)

Approach:

To solve the problem follow the below idea:

Here we use the two variables, one for the vertical distance of the current node from the root and another for the depth of the current node from the root. We use the vertical distance for indexing. If one node with the same vertical distance comes again, we check if the depth of the new node is lower or higher with respect to the current node with the same vertical distance in the map. If the depth of the new node is lower, then we replace it.

Follow the below steps to solve the problem:

1) Create a map of the type <int, pair<int, int>> and two variables d and l to store horizontal and vertical distance from the root respectively
2) Call the function to return the SumOfTopView
3) If the root is equal to the null value then return from the function (Base case)
4) Check if this value of d is not present in the map, then set map[d] equal to {root->data, l}
5) Check if this value of d is already present and its vertical distance is greater than l, then set map[d] equal to {root->data, l}
6) Call this function recursively for (root->left, d-1, l+1, mp) and (root->right, d+1, l+1, mp)
7) return the sum of top view of the binary tree using the map

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Structure of binary tree``struct` `Node {``    ``Node* left;``    ``Node* right;``    ``int` `data;``};` `// function to create a new node``Node* newNode(``int` `data){``    ``Node* temp = ``new` `Node();``    ``temp->left = temp->right = NULL;``    ``temp->data = data;``    ``return` `temp;``}` `// function to fill the map``void` `fillMap(Node* root, ``int` `d, ``int` `l, map<``int``, pair<``int``, ``int``> >& m){``    ``if` `(root == NULL) ``return``;` `    ``if` `(m.count(d) == 0)``        ``m[d] = make_pair(root->data, l);``    ``else` `if``(m[d].second > l)``        ``m[d] = make_pair(root->data, l);``    ` `    ``fillMap(root->left, d - 1, l + 1, m);``    ``fillMap(root->right, d + 1, l + 1, m);``}` `// function return the SumofTopView of``// the given binary tree``int` `SumOfTopView(Node* root){``    ``// map to store the pair of node value and its level``    ``// with respect to the vertical distance from root.``    ``map<``int``, pair<``int``, ``int``> > m;` `    ``// Initializing map``    ``fillMap(root, 0, 0, m);` `    ``int` `ans = 0;``    ``for``(``auto` `i : m){``        ``ans += i.second.first;``    ``}``    ``return` `ans;``}``// Driver code``int` `main(){``    ``Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->right = newNode(4);``    ``root->left->right->right = newNode(5);``    ``root->left->right->right->right = newNode(6);``    ``cout<

## Java

 `import` `java.util.*;` `// structure of binary tree node``class` `Node {``  ``int` `data;``  ``Node left, right;``  ``Node(``int` `data) {``    ``this``.data = data;``    ``left = ``null``;``    ``right = ``null``;``  ``}``}` `// pair class``class` `Pair {``  ``int` `first, second;``  ``Pair(``int` `first, ``int` `second) {``    ``this``.first = first;``    ``this``.second = second;``  ``}``}` `class` `Solution {` `  ``// function to fill the map``  ``public` `void` `fillMap(Node root, ``int` `d, ``int` `l, Map m) {``    ``if` `(root == ``null``)``      ``return``;` `    ``if` `(m.get(d) == ``null``) {``      ``m.put(d, ``new` `Pair(root.data, l));``    ``} ``else` `if` `(m.get(d).second > ``1``) {``      ``m.put(d, ``new` `Pair(root.data, l));``    ``}` `    ``fillMap(root.left, d-``1``, l+``1``, m);``    ``fillMap(root.right, d+``1``, l+``1``, m);``  ``}` `  ``// function returns the sumofTopView of the given binary tree``  ``public` `int` `sumOfTopView(Node root) {``    ``// map to store the pair of node value and its level``    ``// with respect to the vertical distance from root``    ``Map m = ``new` `TreeMap<>();` `    ``// initializing map``    ``fillMap(root, ``0``, ``0``, m);` `    ``int` `ans = ``0``;``    ``for``(Integer x: m.keySet()) {``      ``ans += m.get(x).first;``    ``}` `    ``return` `ans;``  ``}` `  ``// driver program to test above functions``  ``public` `static` `void` `main(String[] args) {``    ``Node root = ``new` `Node(``1``);``    ``root.left = ``new` `Node(``2``);``    ``root.right = ``new` `Node(``3``);``    ``root.left.right = ``new` `Node(``4``);``    ``root.left.right.right = ``new` `Node(``5``);``    ``root.left.right.right.right = ``new` `Node(``6``);``    ``Solution s = ``new` `Solution();``    ``System.out.println(s.sumOfTopView(root));``  ``}``}`

## Python

 `# Python program for the above approach``# structure of binary tree node``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None``    ` `# pair class``class` `pair:``    ``def` `__init__(``self``, first, second):``        ``self``.first ``=` `first``        ``self``.second ``=` `second``    ` `# function to create a new node``def` `newNode(data):``    ``return` `Node(data)`  `# function to fill the map``def` `fillMap(root, d, l, m):``    ``if``(root ``is` `None``):``        ``return``    ` `    ``if``(m.get(d) ``is` `None``):``        ``m[d] ``=` `pair(root.data, l)``    ``elif``(m.get(d).second > ``1``):``        ``m[d] ``=` `pair(root.data, l)``    ` `    ``fillMap(root.left, d``-``1``, l``+``1``, m)``    ``fillMap(root.right, d``+``1``, l``+``1``, m)``    `  `# function returns the sumofTopView of``# the given binary tree``def` `SumOfTopView(root):``    ``# map to store the pair of node value and its level``    ``# with respect to the vertical distance from root``    ``m ``=` `{}``    ` `    ``# initializing map``    ``fillMap(root, ``0``, ``0``, m)``    ` `    ``ans ``=` `0``    ``for` `x ``in` `m:``        ``ans ``+``=` `m[x].first``    ` `    ``return` `ans` `# driver program to test above functions``root ``=` `newNode(``1``)``root.left ``=` `newNode(``2``)``root.right ``=` `newNode(``3``)``root.left.right ``=` `newNode(``4``)``root.left.right.right ``=` `newNode(``5``)``root.left.right.right.right ``=` `newNode(``6``)``print``(SumOfTopView(root))`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``using` `System.Collections;``using` `System.Linq;` `// Structure of binary tree``class` `Node {``  ``public` `Node left;``  ``public` `Node right;``  ``public` `int` `data;` `  ``public` `Node(``int` `val){``    ``data = val;``    ``left = ``null``;``    ``right = ``null``;``  ``}``}` `class` `HelloWorld {` `  ``// function to fill the map``  ``public` `static` `void` `fillMap(Node root, ``int` `d, ``int` `l, IDictionary<``int``, KeyValuePair<``int``,``int``> > m){``    ``if` `(root == ``null``) ``return``;` `    ``if` `(m.ContainsKey(d) == ``false``)``      ``m.Add(``new` `KeyValuePair<``int``, KeyValuePair<``int``,``int``>>(d, ``new` `KeyValuePair<``int``, ``int``>(root.data, l)));``    ``else` `if``(m[d].Value > 1){``      ``m[d] = ``new` `KeyValuePair<``int``,``int``>(root.data, l);``    ``}` `    ``fillMap(root.left, d - 1, l + 1, m);``    ``fillMap(root.right, d + 1, l + 1, m);``  ``}` `  ``// function return the SumofTopView of``  ``// the given binary tree``  ``public` `static` `int` `SumOfTopView(Node root)``  ``{` `    ``// map to store the pair of node value and its level``    ``// with respect to the vertical distance from root.``    ``IDictionary<``int``, KeyValuePair<``int``,``int``> > m = ``new` `Dictionary<``int``,  KeyValuePair<``int``,``int``>>();`  `    ``// Initializing map``    ``fillMap(root, 0, 0, m);` `    ``int` `ans = 0;``    ``foreach``(KeyValuePair<``int``, KeyValuePair<``int``,``int``>> i ``in` `m)``    ``{``      ``ans += i.Value.Key;``      ``// do something with entry.Value or entry.Key``    ``}` `    ``return` `ans;``  ``}` `  ``static` `void` `Main() {``    ``Node root = ``new` `Node(1);``    ``root.left = ``new` `Node(2);``    ``root.right = ``new` `Node(3);``    ``root.left.right = ``new` `Node(4);``    ``root.left.right.right = ``new` `Node(5);``    ``root.left.right.right.right = ``new` `Node(6);``    ``Console.WriteLine(SumOfTopView(root));``  ``}``}` `// The code is contributed by Nidhi goel.`

## Javascript

 `// JavaScript program for the above approach``// structure of binary tree node``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 to create a new node``function` `newNode(data){``    ``return` `new` `Node(data);``}` `// functiont o fill the map``function` `fillMap(root, d, l, m){``    ``if``(root == ``null``) ``return``;``    ` `    ``if``(m.has(d) == ``false``)``        ``m.set(d, ``new` `pair(root.data, l));``    ``else` `if``(m.get(d).second > 1)``        ``m.set(d, ``new` `pair(root.data, l));``    ` `    ``fillMap(root.left, d-1, l+1, m);``    ``fillMap(root.right, d+1, l+1, m);``}` `// function return the SumOfTopView of``// the given binary tree``function` `SumOfTopView(root){``    ``// map to store the pair of node value and its level``    ``// with respect to the vertical distance from root``    ``let m = ``new` `Map();``    ` `    ``// initializing map``    ``fillMap(root, 0, 0, m);``    ` `    ``let ans = 0;``    ``m.forEach(``function``(value, key){``        ``ans += value.first;``    ``})``    ``return` `ans;``}` `// driver code``let root = newNode(1);``root.left = newNode(2);``root.right = newNode(3);``root.left.right = newNode(4);``root.left.right.right = newNode(5);``root.left.right.right.right = newNode(6);``document.write(SumOfTopView(root));` `// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)`

Output

`12`

Time complexity: O(N * log(N)), where N is the number of nodes in the given binary tree.
Auxiliary Space: O(N)