# Print 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, print the top view of it. The output nodes can be printed in any order. Expected time complexity is O(n)

A node x is there in output if x is the topmost node at its horizontal distance. Horizontal distance of left child of a node x is equal to horizontal distance of x minus 1, and that of right child is horizontal distance of x plus 1.

```       1
/     \
2       3
/  \    / \
4    5  6   7
Top view of the above binary tree is
4 2 1 3 7

1
/   \
2       3
\
4
\
5
\
6
Top view of the above binary tree is
2 1 3 6```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to do something similar to vertical Order Traversal. Like vertical Order Traversal, we need 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. Hashing is used to check if a node at given horizontal distance is seen or not.

## C++

 `// C++ program to print top ` `// view of binary tree ` ` `  `#include   ` `#include  ` `#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 should print the topView of ` `// the binary tree ` `void` `topview(Node* root) ` `{ ` `    ``if``(root==NULL) ` `       ``return``; ` `     ``queueq; ` `     ``map<``int``,``int``> m;  ` `     ``int` `hd=0; ` `     ``root->hd=hd; ` `      `  `     ``// push node and horizontal distance to queue ` `    ``q.push(root); ` `     `  `    ``cout<< ``"The top view of the tree is : \n"``; ` `     `  `    ``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; ` `        ``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(); ` `       `  `    ``} ` `     `  `     `  `     `  `     ``for``(``auto` `i=m.begin();i!=m.end();i++) ` `    ``{ ` `        ``cout<second<<``" "``; ` `    ``} ` `     `  `} ` `  `  `// Driver Program to test above functions ` `int` `main() ` `{ ` `    ``/* Create following Binary Tree  ` `            ``1  ` `        ``/ \  ` `        ``2 3  ` `        ``\  ` `            ``4  ` `            ``\  ` `            ``5  ` `            ``\  ` `                ``6*/` `   ``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<<``"Following are nodes in top view of Binary Tree\n"``;  ` `    ``topview(root); ` `    ``return` `0; ` `} ` `/* This code is contributed by Niteesh Kumar */`

## Java

 `// Java program to print top ` `// view of binary tree ` `import` `java.util.Queue; ` `import` `java.util.TreeMap; ` `import` `java.util.LinkedList; ` `import` `java.util.Map; ` `import` `java.util.Map.Entry; ` ` `  `// class to create a node ` `class` `Node { ` `    ``int` `data; ` `    ``Node left, right; ` ` `  `    ``public` `Node(``int` `data) { ` `        ``this``.data = data; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `// class of binary tree ` `class` `BinaryTree { ` `    ``Node root; ` ` `  `    ``public` `BinaryTree() { ` `        ``root = ``null``; ` `    ``} ` `     `  `    ``// function should print the topView of ` `    ``// the binary tree ` `    ``private` `void` `TopView(Node root) { ` `        ``class` `QueueObj { ` `            ``Node node; ` `            ``int` `hd; ` ` `  `            ``QueueObj(Node node, ``int` `hd) { ` `                ``this``.node = node; ` `                ``this``.hd = hd; ` `            ``} ` `        ``} ` `        ``Queue q = ``new` `LinkedList(); ` `        ``Map topViewMap = ``new` `TreeMap(); ` ` `  `        ``if` `(root == ``null``) { ` `            ``return``; ` `        ``} ``else` `{ ` `            ``q.add(``new` `QueueObj(root, ``0``)); ` `        ``} ` ` `  `        ``System.out.println(``"The top view of the tree is : "``); ` `         `  `        ``// count function returns 1 if the container  ` `        ``// contains an element whose key is equivalent  ` `        ``// to hd, or returns zero otherwise. ` `        ``while` `(!q.isEmpty()) { ` `            ``QueueObj tmpNode = q.poll(); ` `            ``if` `(!topViewMap.containsKey(tmpNode.hd)) { ` `                ``topViewMap.put(tmpNode.hd, tmpNode.node); ` `            ``} ` ` `  `            ``if` `(tmpNode.node.left != ``null``) { ` `                ``q.add(``new` `QueueObj(tmpNode.node.left, tmpNode.hd - ``1``)); ` `            ``} ` `            ``if` `(tmpNode.node.right != ``null``) { ` `                ``q.add(``new` `QueueObj(tmpNode.node.right, tmpNode.hd + ``1``)); ` `            ``} ` ` `  `        ``} ` `        ``for` `(Entry entry : topViewMap.entrySet()) { ` `            ``System.out.print(entry.getValue().data); ` `        ``} ` `    ``} ` `     `  `    ``// Driver Program to test above functions ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{  ` `        ``/* Create following Binary Tree  ` `            ``1  ` `        ``/ \  ` `        ``2 3  ` `        ``\  ` `            ``4  ` `            ``\  ` `            ``5  ` `            ``\  ` `                ``6*/` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``1``); ` `        ``tree.root.left = ``new` `Node(``2``); ` `        ``tree.root.right = ``new` `Node(``3``); ` `        ``tree.root.left.right = ``new` `Node(``4``); ` `        ``tree.root.left.right.right = ``new` `Node(``5``); ` `        ``tree.root.left.right.right.right = ``new` `Node(``6``); ` `        ``System.out.println(``"Following are nodes in top view of Binary Tree"``);  ` `        ``tree.TopView(tree.root);  ` `    ``}  ` `     `  `} `

## Python3

 `# Python3 program to print top  ` `# view of binary tree ` ` `  `# 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` `        ``self``.hd ``=` `0` ` `  `# function should print the topView  ` `# of the binary tree  ` `def` `topview(root) : ` ` `  `    ``if``(root ``=``=` `None``) : ` `        ``return` `    ``q ``=` `[] ` `    ``m ``=` `dict``() ` `    ``hd ``=` `0` `    ``root.hd ``=` `hd  ` ` `  `    ``# push node and horizontal ` `    ``# distance to queue  ` `    ``q.append(root)  ` ` `  `    ``while``(``len``(q)) : ` `        ``root ``=` `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` `hd ``not` `in` `m: ` `            ``m[hd] ``=` `root.data  ` `        ``if``(root.left) :          ` `            ``root.left.hd ``=` `hd ``-` `1` `            ``q.append(root.left)  ` `         `  `        ``if``(root.right):          ` `            ``root.right.hd ``=` `hd ``+` `1` `            ``q.append(root.right)  ` `         `  `        ``q.pop(``0``) ` `    ``for` `i ``in` `sorted` `(m): ` `        ``print``(m[i], end ``=` `"")  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``""" Create following Binary Tree  ` `            ``1  ` `        ``/ \  ` `        ``2 3  ` `        ``\  ` `            ``4  ` `            ``\  ` `            ``5  ` `            ``\  ` `                ``6*"""` `    ``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``(``"Following are nodes in top"``,  ` `          ``"view of Binary Tree"``)  ` `    ``topview(root) ` ` `  `# This code is contributed by ` `# Shubham Singh(SHUBHAMSINGH10) `

Output:

```Following are nodes in top view of Binary Tree
2136```

Another approach:

This approach does not require a queue. Here we use the two variables, one for vertical distance of 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 depth of new node is lower or higher with respect to the current node with same vertical distance in the map. If depth of new node is lower, then we replace it.

## C++

 `#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` `key){ ` `    ``Node* node=``new` `Node(); ` `    ``node->left = node->right = NULL; ` `    ``node->data=key; ` `    ``return` `node; ` `} ` ` `  `// 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 should print the topView of ` `// the binary tree ` `void` `topView(``struct` `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; ` ` `  `    ``//fillmap(root,vectical_distance_from_root,level_of_node,map) ` `    ``fillMap(root,0,0,m); ` ` `  `    ``for``(``auto` `it=m.begin();it!=m.end();it++){ ` `        ``cout << it->second.first << ``" "``; ` `    ``} ` `} ` `// Driver Program to test above functions ` `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<<``"Following are nodes in top view of Binary Tree\n"``;  ` `    ``topView(root); ` `    ``return` `0; ` `} ` `/* This code is contributed by Akash Debnath */`

## Python3

 `# 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` ` `  `# function to fill the map ` `def` `fillMap(root, d, l, m): ` `    ``if``(root ``=``=` `None``): ` `        ``return` `     `  `    ``if` `d ``not` `in` `m: ` `        ``m[d] ``=` `[root.data,l] ` `    ``elif``(m[d][``1``] > l): ` `        ``m[d] ``=` `[root.data,l] ` `    ``fillMap(root.left, d ``-` `1``, l ``+` `1``, m) ` `    ``fillMap(root.right, d ``+` `1``, l ``+` `1``, m) ` ` `  `# function should prthe topView of ` `# the binary tree ` `def` `topView(root): ` `     `  `    ``# map to store the pair of node value and its level  ` `    ``# with respect to the vertical distance from root.  ` `    ``m ``=` `{} ` `     `  `    ``fillMap(root, ``0``, ``0``, m) ` `    ``for` `it ``in` `sorted` `(m.keys()): ` `        ``print``(m[it][``0``], end ``=` `" "``) ` `     `  `# Driver Code ` `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``(``"Following are nodes in top view of Binary Tree"``) ` `topView(root) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

Output:

```Following are nodes in top view of Binary Tree
2 1 3 6```

This approach is contributed by Akash Debnath

Time Complexity of the above implementation is O(n) where n is the number of nodes in the given binary tree. The assumption here is that add() and contains() methods of HashSet work in O(1) time.