Top View Of Binary Tree

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Problem statement

You are given a Binary Tree of 'n' nodes.


The Top view of the binary tree is the set of nodes visible when we see the tree from the top.


Find the top view of the given binary tree, from left to right.


Example :
Input: Let the binary tree be:

Example

Output: [10, 4, 2, 1, 3, 6]

Explanation: Consider the vertical lines in the figure. The top view contains the topmost node from each vertical line.
Detailed explanation ( Input/output format, Notes, Images )
Input format :
The first line of input contains elements in the level order form for the first binary tree. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.
Input format explanation:
The level order input for the tree depicted in the below image would be 

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level, and so on.
The input ends when all nodes at the last level are null (-1).


Output Format:
Print the vector/list of all the elements of the top view of the given tree from left to right.


Note :
You do not need to print anything; it has already been taken care of. Just implement the given function.
Sample Input 1:
1 2 3 4 5 -1 6 -1 7 -1 -1 8 -1 9 -1 -1 11 10 -1 -1 -1 -1 -1


Sample Output 1:
10 4 2 1 3 6


Explanation of Sample Output 1:
The binary tree is:

Example

Consider the vertical lines in the figure. The top view contains the topmost node from each vertical line.

In test case 1,


Sample Input 2:
1 2 3 4 5 6 7 8 9 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1


Sample Output 2:
8 4 2 1 3 7


Explanation of Sample Output 2:
The binary tree is:

Example

From left to right, the top view of the tree will be [8,4,2,1,3,7], where 9, 5 and 6 will be hidden when we see from the top of the tree.


Expected time complexity :
The expected time complexity is O(n * log(n)).


Constraints:
1 <= 'n' <= 10000
1 <= data in any node <= 10 ^ 6

Time limit: 1 sec
Hint

How will a node be hidden by another node?

Approaches (2)
Using Pre-Order Traversal

As we know that all three traversals, i.e. pre-order, in-order and post-order, visit the tree node at once. We can use any of them. Here we are going to use pre-order traversal for the explanation. So while traversing in the pre-order traversal, we will keep track of horizontal distance of the node which is going to be visited from the root node, and we also keep track of the vertical level of that node (where the vertical level of root node would be ‘0’ and its children would be ‘1’ and so on.... ). We will be using the Map which stores the horizontal distance of the node from the root as the key and value in the map maintains a pair of containing the value of the node and the level of the node. 

 

The steps are as follows:

  1. Make a Map like:
    • map <int, pair<int, int> > visited, where the key of ‘visited’ defines the horizontal distance and value in the map maintains a pair of containing the value of the node and the level of the node.
  2. And now call preOrder function. Here, ‘hDistance’ defines the horizontal distance and level defines the depth of the tree.
applyPreorder(root , hDistance, level,visited) = { applyPreorder(root->left , hDistance-1, level+1,visited) , applyPreorder(root->right , hDistance+1, level+1,visited )}

    3. For every ‘hDistance’ check whether it is visited or not? If it is not visited, then make it visited with the value of node and ‘level’ and if it is already visited, then check if the previous stored ‘level’ is greater than then the current level. If yes, then store the current node because the previous node now is hidden by the current node because the current node has a lesser height.

     4. Once we are done with pre-order, our map contains ‘hDistance’ as the key and value corresponding to each ‘hDistance’ stores the pair of nodes and their level that are visible from the top of the tree at that ‘hDistance’. So iterate over the map and store the value of the node in array/list. Let’s say ‘topView’.

    5. Return the ‘topView’.

Time Complexity

O(n * log(n)), Where 'n' is the number of nodes in the given binary tree.

Since we are traversing at every node and storing the horizontal distance of every node into Map, inserting an element into the map will take O(log n) time. In the worst case (Skewed Trees), there will be 'n' distinct horizontal distance, so it will take O(n * log(n)). While for the 'n' element, push and pop operation take O(1) time for the queue and also inserting and searching into the list take O(1) time. So overall time complexity will be O(n * log(n)).

Space Complexity

O(n), Where 'n' is the number of nodes in the given binary tree.

Since we are storing the top view of the tree, so in the worst case (Skewed Trees), all nodes of the given tree will be the top view elements so, there will be 'n' distinct horizontal distance to be stored and also in that case stack size for recursion will be O(n). So overall space complexity will be O(n).

Code Solution
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Top View Of Binary Tree
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Interview problems

Top View Of Binary Tree | Similar to vertical order traversal | Java|

class Pair{

        int line;

        TreeNode node;

        Pair(int _line, TreeNode _node){

            this.line = _line;

            this.node = _node;

        }

}

public class Solution {

    public static List<Integer> getTopView(TreeNode root) {

        Queue<Pair> q = new LinkedList<>();

        Map<Integer, Integer> map = new TreeMap<>();

        List<Integer> list = new ArrayList<>();

 

        q.offer(new Pair(0, root));

        while(!q.isEmpty()){

            Pair pair = q.poll();

            int vl = pair.line; //vl-> vertical line

            TreeNode temp = pair.node;

 

            if(map.get(vl) == null){

                map.put(vl, temp.data);

            }

 

            if(temp.left != null){

                q.offer(new Pair(vl-1, temp.left));

            }

 

            if(temp.right != null){

                q.offer(new Pair(vl+1, temp.right));

            }

        }

 

        for(Map.Entry<Integer,Integer> entry : map.entrySet()){

            list.add(entry.getValue());

        }

 

        return list;

    }

}

 

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Interview problems

Java solution

class Pair{

    int hd;

    TreeNode node;

    public Pair(int hd,TreeNode node){

        this.hd = hd;

        this.node = node;

    }

}

public class Solution {

    public static List<Integer> getTopView(TreeNode root) {

        // Write your code here.

        Map<Integer,Integer> map = new TreeMap<>();

        List<Integer> ans = new ArrayList<>();

        if(root == null)

        return ans;

        Queue<Pair> q = new LinkedList<>();

        q.add(new Pair(0,root));

        while(!q.isEmpty()){

            Pair p = q.poll();

            TreeNode node = p.node;

            int hd = p.hd;

          map.putIfAbsent(hd, node.data);

          if(node.left != null)

          q.add(new Pair(hd-1,node.left));

          if(node.right!=null)

          q.add(new Pair(hd+1,node.right));

        }

        for(Map.Entry<Integer,Integer> entry: map.entrySet()){

            ans.add(entry.getValue());

        }

        return ans;

    }

}

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Interview problems

STL not working ??

stl is not working in my code like map and queues.

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Interview problems

Top View of Binary Tree using Java

import java.util.*;

 

class TreeNodeWithHorizontalDistance {

    TreeNode node;

    int horizontalDistance;

 

    TreeNodeWithHorizontalDistance(TreeNode node, int horizontalDistance) {

        this.node = node;

        this.horizontalDistance = horizontalDistance;

    }

}

 

public class Solution {

    public static List<Integer> getTopView(TreeNode root) {

        if (root == null) return new ArrayList<>();

 

        List<Integer> topViewNodes = new ArrayList<>();

        Map<Integer, Integer> topViewMap = new TreeMap<>(); // Horizontal distance to node data mapping

        Queue<TreeNodeWithHorizontalDistance> nodeQueue = new LinkedList<>();

 

        nodeQueue.add(new TreeNodeWithHorizontalDistance(root, 0));

 

        while (!nodeQueue.isEmpty()) {

            TreeNodeWithHorizontalDistance currentNodeWithHD = nodeQueue.poll(); // Retrieve the node and horizontal distance

            int horizontalDistance = currentNodeWithHD.horizontalDistance;

            TreeNode currentNode = currentNodeWithHD.node;

 

            if (!topViewMap.containsKey(horizontalDistance)) {

                topViewMap.put(horizontalDistance, currentNode.data);

            }

 

            if (currentNode.left != null) {

                nodeQueue.add(new TreeNodeWithHorizontalDistance(currentNode.left, horizontalDistance - 1));

            }

 

            if (currentNode.right != null) {

                nodeQueue.add(new TreeNodeWithHorizontalDistance(currentNode.right, horizontalDistance + 1));

            }

        }

 

        for (Map.Entry<Integer, Integer> entry : topViewMap.entrySet()) {

            topViewNodes.add(entry.getValue());

        }

 

        return topViewNodes;

    }

}

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Interview problems

✅SIMPLE and easy MAPPING🔥🔥

#include<bits/stdc++.h>

 

vector<int> getTopView(TreeNode<int> *root)

{

 

    vector<int> ans;

    queue<pair<TreeNode<int> *,int>> q;

    q.push({root,0});

    map<int,int> mp;

 

    while(!q.empty()){

        TreeNode<int> * temp=q.front().first;

        int level=q.front().second;

        q.pop();

 

        if(mp.find(level)==mp.end()){

            mp[level]=temp->data;

        }

 

        if(temp->left){

            q.push({temp->left,level-1});

        }

        if(temp->right){

            q.push({temp->right,level+1});

        }

        

        

    }

    for(auto &it:mp){

        ans.push_back(it.second);

    }

 

    return ans;

 

}

 

programming

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Interview problems

C++ Unique Solution without Queue and Level order

#include <map>

#include <vector>

using namespace std;

void code(TreeNode<int> *root,int row,int col , map<int,pair<int,int>> &mp)

{

    if ( root == NULL)

    return ;

    if(mp.find(col)==mp.end())

    {

    mp[col]={row,root->data};

    }

    else{

    if (mp[col].first>row)

      mp[col] = {row, root->data};

    }

 

    code(root->left,row+1,col-1,mp);

    code(root->right,row+1,col+1,mp);

}

vector<int> getTopView(TreeNode<int> *root)

{

    map<int,pair<int,int>>mp;

    vector<int>ans;

    code(root,0,0,mp);

    for(auto  it : mp)

    {

        ans.push_back(it.second.second);

    }

    return ans ;

}

 

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Interview problems

Top View Python

def getTopView(root):

    # Write your code here.

 

    bfs = []

    dicts = defaultdict(list)

    if(root is None):

        return root

    queue = deque([(root, 0)])

    while(queue):

        l = len(queue)

        for i in range(0, l):

            curr, nl = queue.popleft()

            dicts[nl].append(curr.val)

            if(curr.left):

                queue.append((curr.left, nl-1))

            if(curr.right):

                queue.append((curr.right, nl+1))

    # print(dicts)

    dicts1 = sorted(dicts.items(), key=lambda x: x[0])

    for key, value in dicts1:

        bfs.append(value[0])

    return bfs

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Interview problems

top view of binary tree | easy c++ solution of striver dsa a to z sheet

my github : https://github.com/yashswag22

 

#include<bits/stdc++.h>

 

vector<int> getTopView(TreeNode<int> *root)

{

    vector<int>ans;

    if(root == NULL) return ans;

 

    map <int, map<int,vector<int>>> nodes;

 

    // to fulfill this "If two nodes have the same position, then the value of the node that is added first will be the value that is on the left side."

    // we will use level order traversal

 

    queue< pair< TreeNode<int>*, pair<int,int>>> q;

 

    q.push({root,{0,0}});

 

    while(!q.empty()){

        pair< TreeNode<int>*, pair<int,int>> temp = q.front();

        q.pop();

        TreeNode<int>* frontNode = temp.first;

        int hd = temp.second.first; // hd = horizontal distance

        int lvl = temp.second.second; // lvl = level of node 

 

        if(frontNode!= NULL){

            nodes[hd][lvl].push_back(frontNode->data);

        }

 

        if(frontNode->left)

        q.push({frontNode ->left,{hd-1,lvl+1}});

        if(frontNode->right)

        q.push({frontNode ->right,{hd+1,lvl+1}});

    }

 

    for(auto i : nodes){ // it give <int, map<>>

        for(auto j : i.second){ // it give <int, vector<>>

           for(auto k : j.second) { // it will iterate the vector

                ans.push_back(k);

                break;

           }

            break;

        }

    }

 

    return ans;

}

tutorial

100daysofcode

Top View Of Binary Tree

Top View of the Binary Tree

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Interview problems

C++ || using map || level order traversal


#include <bits/stdc++.h>
vector<int> getTopView(TreeNode<int> *root) {
    vector<int> ans;
    if (root == NULL) {
        return ans;
    }
    
    // Mapping of horizontal distance with node data
    map<int, int> topNode;
    // Queue for level order traversal, stores nodes with their horizontal distance
    queue<pair<TreeNode<int>*, int>> q;
    q.push(make_pair(root, 0)); // The root has a horizontal distance of 0
    
    while (!q.empty()) {
        pair<TreeNode<int>*, int> temp = q.front();
        q.pop();
        TreeNode<int>* frontNode = temp.first;
        int hd = temp.second;
        
        // If a value is not yet present for a horizontal distance, add it
        if (topNode.find(hd) == topNode.end()) {
            topNode[hd] = frontNode->data;
        }
        
        if (frontNode->left) {
            q.push(make_pair(frontNode->left, hd - 1));
        }
        
        if (frontNode->right) {
            q.push(make_pair(frontNode->right, hd + 1));
        }
    }
    
    for (auto i: topNode) {
        ans.push_back(i.second);
    }
    
    return ans;
}
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Interview problems

T.C: O(N) solution without using map

/*********************************************

 Following is the TreeNode class structure

 class TreeNode {
     int data;
     TreeNode left;
     TreeNode right;

     TreeNode(int data) {
         this.data = data;
         this.left = null;
         this.right = null;
     }
 }

 *********************************************/

import java.util.*;


public class Solution {
    static class Pair{
        TreeNode node;
        int coord;

        public Pair(TreeNode node, int coord){
            this.node = node;
            this.coord = coord;
        }
    }
    public static List<Integer> getTopView(TreeNode root) {
        // Write your code here.
        int min = 1, max = 0;
        List<Integer> a = new ArrayList<>(), b = new ArrayList<>();
        Queue<Pair> queue = new LinkedList<>();
        queue.add(new Pair(root, 0));

        while(!queue.isEmpty()){
            Pair pair = queue.poll();
            int coord = pair.coord;
            TreeNode left = pair.node.left, right = pair.node.right;
            if (coord < min){
                a.add(pair.node.data);
                min = coord;
            }
            if (coord > max){
                b.add(pair.node.data);
                max = coord;
            }
            if (left != null)
                queue.add(new Pair(left, coord-1));
            if (right != null)
                queue.add(new Pair(right, coord+1));
        }
        Collections.reverse(a);
        a.addAll(b);
        return a;
    }
}
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