# Sum of nodes in bottom view of Binary Tree

Given a binary tree, the task is to print the sum of nodes in the bottom view of the given Binary Tree. The bottom view of a binary tree is the set of nodes visible when the tree is viewed from the bottom.
Examples:

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

Output : 20

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

Output : 17

```

Approach: The idea is to use a queue.

1. Put tree nodes in a queue for the level order traversal
2. Start with the horizontal distance(hd) 0 of the root node, keep on adding left child to queue along with the horizontal distance as hd-1 and right child as hd+1.
3. Use a map to store the hd value(as key) and the last node(as pair) having the corresponding hd value.
4. Every time, we encounter a new horizontal distance or an existing horizontal distance put the node data for the horizontal distance as key. For the first time it will add to the map, next time it will replace the value. This will make sure that the bottom-most element for that horizontal distance is present in the map.
5. Finally, traverse the map and calculate the sum of all the elements.

Below is the implementation of the above approach:

 `// 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 bottom view of binary tree` `int` `SumOfBottomView(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()) {` `        ``Node* temp = q.front();` `        ``q.pop();`   `        ``hd = temp->hd;`   `        ``// Put the dequeued tree node to Map` `        ``// having key as horizontal distance. Every` `        ``// time we find a node having same horizontal` `        ``// distance we need to replace the data in` `        ``// the map.` `        ``m[hd] = temp->data;`   `        ``if` `(temp->left) {` `            ``temp->left->hd = hd - 1;` `            ``q.push(temp->left);` `        ``}` `        ``if` `(temp->right) {` `            ``temp->right->hd = hd + 1;` `            ``q.push(temp->right);` `        ``}` `    ``}`   `    ``map<``int``, ``int``>::iterator itr;`   `    ``// Sum up all the nodes in bottom view of the tree` `    ``for` `(itr = m.begin(); itr != m.end(); itr++)` `        ``sum += itr->second;`   `    ``return` `sum;` `}`   `// Driver Code` `int` `main()` `{` `    ``Node* root = newNode(20);` `    ``root->left = newNode(8);` `    ``root->right = newNode(22);` `    ``root->left->left = newNode(5);` `    ``root->left->right = newNode(3);` `    ``root->right->left = newNode(4);` `    ``root->right->right = newNode(25);` `    ``root->left->right->left = newNode(10);` `    ``root->left->right->right = newNode(14);`   `    ``cout << SumOfBottomView(root);`   `    ``return` `0;` `}`

 `# Python3 implementation of the approach `   `class` `Node: ` `    `  `    ``def` `__init__(``self``, data):` `        ``self``.data ``=` `data` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `        ``self``.hd ``=` `None`   `# Function that returns the Sum of ` `# nodes in bottom view of binary tree ` `def` `SumOfBottomView(root): `   `    ``if` `root ``=``=` `None``:` `        ``return` `0` `    `  `    ``q ``=` `[] ` `    ``m ``=` `{}` `    ``hd, ``Sum` `=` `0``, ``0` `    ``root.hd ``=` `hd `   `    ``# Push node and horizontal ` `    ``# distance to queue ` `    ``q.append(root) `   `    ``while` `len``(q) > ``0``: ` `        ``temp ``=` `q.pop(``0``) ` `        ``hd ``=` `temp.hd `   `        ``# Put the dequeued tree node to Map ` `        ``# having key as horizontal distance. ` `        ``# Every time we find a node having ` `        ``# same horizontal distance we need ` `        ``# to replace the data in the map. ` `        ``m[hd] ``=` `temp.data `   `        ``if` `temp.left !``=` `None``: ` `            ``temp.left.hd ``=` `hd ``-` `1` `            ``q.append(temp.left) ` `        `  `        ``if` `temp.right !``=` `None``: ` `            ``temp.right.hd ``=` `hd ``+` `1` `            ``q.append(temp.right) `   `    ``# Sum up all the nodes in bottom ` `    ``# view of the tree ` `    ``for` `itr ``in` `m:` `        ``Sum` `+``=` `m[itr] `   `    ``return` `Sum`   `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: `   `    ``root ``=` `Node(``20``) ` `    ``root.left ``=` `Node(``8``) ` `    ``root.right ``=` `Node(``22``) ` `    ``root.left.left ``=` `Node(``5``) ` `    ``root.left.right ``=` `Node(``3``) ` `    ``root.right.left ``=` `Node(``4``) ` `    ``root.right.right ``=` `Node(``25``) ` `    ``root.left.right.left ``=` `Node(``10``) ` `    ``root.left.right.right ``=` `Node(``14``) `   `    ``print``(SumOfBottomView(root)) ` `    `  `# This code is contributed by Rituraj Jain`

 `// C# implementation of ` `// the above approach` `using` `System;` `using` `System.Collections; ` `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.data = key;` `  ``node.left = ``null``;` `  ``node.right = ``null``;` `  ``return` `node;` `}` ` `  `// Function that returns the sum of` `// nodes in bottom view of binary tree` `static` `int` `SumOfBottomView(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) ` `  ``{` `    ``Node temp = (Node)q.Peek();` `    ``q.Dequeue();` `    ``hd = temp.hd;`   `    ``// Put the dequeued tree node ` `    ``// to Map having key as horizontal ` `    ``// distance. Every time we find a ` `    ``// node having same horizontal distance ` `    ``// we need to replace the data in` `    ``// the map.` `    ``m[hd] = temp.data;`   `    ``if` `(temp.left != ``null``) ` `    ``{` `      ``temp.left.hd = hd - 1;` `      ``q.Enqueue(temp.left);` `    ``}` `    ``if` `(temp.right != ``null``) ` `    ``{` `      ``temp.right.hd = hd + 1;` `      ``q.Enqueue(temp.right);` `    ``}` `  ``}    ` `  `  `  ``// Sum up all the nodes in ` `  ``// bottom view of the tree    ` `  ``foreach``(KeyValuePair<``int``,` `                       ``int``> itr ``in` `m)` `  ``{` `    ``sum += itr.Value;` `  ``}` `  ``return` `sum;` `}` ` `  `// Driver code ` `public` `static` `void` `Main(``string``[] args) ` `{` `  ``Node root = newNode(20);` `  ``root.left = newNode(8);` `  ``root.right = newNode(22);` `  ``root.left.left = newNode(5);` `  ``root.left.right = newNode(3);` `  ``root.right.left = newNode(4);` `  ``root.right.right = newNode(25);` `  ``root.left.right.left = newNode(10);` `  ``root.left.right.right = newNode(14);` `  ``Console.Write(SumOfBottomView(root));` `}` `}`   `// This code is contributed by rutvik_56`

Output:
```58

```

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Improved By : rituraj_jain, rutvik_56

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