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2 days ago+ 0 comments

Time Complexity :- O(n) ------ HashMap

void topView(Node * root) {
        map<int,int>i;
        queue<pair<Node*,int>>q;q.push({root,500});
        while(!q.empty()){
            pair<Node*,int> t=q.front();q.pop();
            if(t.first->left)q.push({t.first->left,t.second-1});
            if(t.first->right)q.push({t.first->right,t.second+1});
            if(!i[t.second])i[t.second]=t.first->data;
        }
        for(auto k:i)cout<<k.second<<" ";
    }

1 week ago+ 0 comments

bloc_lst = []
def blocked_list(root):
    if root is None:
        return
    
    if root.left:
        if root.left.right:
            #print(root.left.right.info,end=" ")
            bloc_lst.append(root.left.right.info)
    
    if root.right:
        if root.right.left:
            #print(root.right.left.info,end=" ")
            bloc_lst.append(root.right.left.info)
    
    blocked_list(root.left)
    blocked_list(root.right)
    
from collections import defaultdict

def topView(root):
    if not root:
        return
    
    blocked_list(root)
    
    blocked_nodes = bloc_lst

    # Create a dictionary to store nodes at each horizontal distance from the root
    hd_map = defaultdict(list)

    # Create a queue for level-order traversal
    queue = [(root, 0)]  # Each element is a tuple (node, horizontal distance)

    while queue:
        node, hd = queue.pop(0)

        # Check if the node is not blocked
        if node.info not in blocked_nodes:
            hd_map[hd].append(node.info)

        if node.left:
            queue.append((node.left, hd - 1))
        if node.right:
            queue.append((node.right, hd + 1))

    # Print the top view nodes in order of horizontal distance
    for hd in sorted(hd_map):
        print(hd_map[hd][0], end=" ")

3 weeks ago+ 1 comment

is this right?

include

using namespace std;

struct Node { int data; Node* left; Node* right; };

void topView(Node* root) { if (root == nullptr) { return; }

// Create a map to store the horizontal distance and the corresponding node data
map<int, int> verticalMap;

// Create a queue for level-order traversal
queue<pair<Node*, int>> levelOrder;

// Initialize the queue with the root and horizontal distance 0
levelOrder.push({root, 0});

while (!levelOrder.empty()) {
    Node* current = levelOrder.front().first;
    int horizontalDistance = levelOrder.front().second;
    levelOrder.pop();

    // If this horizontal distance is not in the map, add it to the map
    if (verticalMap.find(horizontalDistance) == verticalMap.end()) {
        verticalMap[horizontalDistance] = current->data;
    }

    // Enqueue the left and right children with updated horizontal distances
    if (current->left != nullptr) {
        levelOrder.push({current->left, horizontalDistance - 1});
    }
    if (current->right != nullptr) {
        levelOrder.push({current->right, horizontalDistance + 1});
    }
}

// Print the top view from left to right
for (const auto& entry : verticalMap) {
    cout << entry.second << ' ';
}

}

int main() { // Create a sample binary tree and call topView Node* root = new Node{1, new Node{2, nullptr, nullptr}, new Node{3, nullptr, nullptr}}; cout << "Top View of Binary Tree: "; topView(root);

In this version of the code, I've used a map to store the nodes seen at each horizontal distance, ensuring that only the topmost node at each distance is recorded. This simplifies the code and ensures that it works correctly for various binary tree structures. cout << endl;

return 0;

}

2 months ago+ 0 comments

runs in O(n) time, not O(nlog n), no need to search for horizontal level in a map, the checking is done in constant time with the leftside and rightside vectors

void topView(Node * root) {
        queue<pair<Node*, int>> levelOrder;
        vector<int> leftside = {root->data};
        vector<int> rightside;
        levelOrder.push({root, 0});
        while (!levelOrder.empty()) {
            Node* temp = levelOrder.front().first;
            int currentHorizontal = levelOrder.front().second;
            if (currentHorizontal <=0 and abs(currentHorizontal) == leftside.size()) {
                leftside.push_back(temp->data);
            } else if (currentHorizontal > 0 and currentHorizontal == rightside.size()+1) {
                rightside.push_back(temp->data);
            }
            if (temp->left != NULL) {
                levelOrder.push({temp->left, currentHorizontal-1});
            }
            if (temp->right != NULL) {
                levelOrder.push({temp->right, currentHorizontal+1});
            }
            levelOrder.pop();
        }
        for (int i=leftside.size()-1; i >= 0; i--) {
            cout << leftside[i] << ' ';
        }
        for (int i=0; i < rightside.size(); i++) {
            cout << rightside[i] << ' ';
        }
    }

2 months ago+ 0 comments

python 3                                                                                                                       
def topView(root):                                                                                             
    queue=[]                                                                                                       
    top_view={}                                                                                                   
    queue.append([root,0])                                                                              
    while(queue):                                                                                               
        node,level=queue.pop(0)  # to remove front element                       
        if level not in top_view:                                                                           
            top_view[level]=node.info                                                                                                                     
        if node.left:
            node.left.level=level-1
            queue.append((node.left, level - 1))
        if node.right:
            node.right.level=level+1
            queue.append((node.right, level + 1))
    print(top_view)
    for level in sorted(top_view):
        print(top_view[level],end=" ")

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