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Sum of nodes in top view of binary tree
• Difficulty Level : Medium
• Last Updated : 02 Mar, 2021

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`
Output:
`12`

Time Complexity: O(N)

Auxiliary Space: O(N)

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