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BT_leftview.cpp
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#include <bits/stdc++.h>
using namespace std;
// Tree Node structure
struct Node {
int data;
Node* left;
Node* right;
};
vector<int> leftView(struct Node *root);
// Utility function to create a new Tree Node
Node* newNode(int val) {
Node* temp = new Node;
temp->data = val;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to build a tree from a given string
Node* buildTree(string str) {
// Corner case: if the input string is empty or starts with 'N', return NULL
if(str.length() == 0 || str[0] == 'N')
return NULL;
// Vector to store split string values
vector<string> ip;
istringstream iss(str);
for(string str; iss >> str; )
ip.push_back(str);
// Create the root of the tree
Node* root = newNode(stoi(ip[0]));
// Queue to store nodes for level order insertion
queue<Node*> queue;
queue.push(root);
// Iterate over the string values to build the tree
int i = 1;
while(!queue.empty() && i < ip.size()) {
// Get and remove the front of the queue
Node* currNode = queue.front();
queue.pop();
// Get the current node's value from the string
string currVal = ip[i];
// If the left child is not null
if(currVal != "N") {
// Create the left child for the current node
currNode->left = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->left);
}
// For the right child
i++;
if(i >= ip.size())
break;
currVal = ip[i];
// If the right child is not null
if(currVal != "N") {
// Create the right child for the current node
currNode->right = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->right);
}
i++;
}
return root;
}
int main() {
int t;
scanf("%d ", &t);
while(t--) {
string s;
getline(cin, s);
Node* root = buildTree(s);
vector<int> vec = leftView(root);
for(int x : vec)
cout << x << " ";
cout << endl;
}
return 0;
}
/* A binary tree node
struct Node
{
int data;
struct Node* left;
struct Node* right;
Node(int x){
data = x;
left = right = NULL;
}
};
*/
// Function to return a list containing elements of the left view of the binary tree
void solve(Node* root, vector<int> &ans, int level) {
// Base case: if the node is null, return
if(root == NULL)
return;
// If we entered into a new level, add the node's data to the answer
if(level == ans.size())
ans.push_back(root->data);
// Recursively call for the left and right subtrees
solve(root->left, ans, level + 1);
solve(root->right, ans, level + 1);
}
vector<int> leftView(Node *root) {
vector<int> ans;
solve(root, ans, 0);
return ans;
}