Defines the `largestValues` function, which returns a vector containing the largest value from each row in the binary tree.
2 : Queue Initialization
queue<TreeNode*> q;
Initializes a queue to store nodes for the breadth-first traversal of the tree.
3 : Result Vector Initialization
vector<int> res;
Creates a vector `res` to store the largest value for each level of the tree.
4 : Check for Empty Tree
if(!root) return res;
Checks if the tree is empty. If it is, returns the empty result vector.
5 : Push Root to Queue
q.push(root);
Pushes the root node of the tree into the queue to begin the breadth-first traversal.
6 : Initialize Max Value
int mx;
Declares a variable `mx` to store the largest value at each level of the tree during traversal.
7 : BFS Loop Start
while(!q.empty()) {
Starts the outer while loop, which will run as long as there are nodes in the queue.
8 : Level Size Calculation
int sz = q.size();
Calculates the number of nodes at the current level by checking the size of the queue.
9 : Initialize Max Value for Level
mx = INT_MIN;
Initializes `mx` to the minimum possible integer value before comparing it with node values at the current level.
10 : Level Traversal Loop
while(sz-->0) {
Begins an inner loop to process all nodes at the current level.
11 : Process Current Node
root = q.front();
Retrieves the front node from the queue for processing.
12 : Pop Node from Queue
q.pop();
Removes the front node from the queue after processing it.
13 : Update Max Value
mx = max(mx, root->val);
Updates the `mx` variable to hold the maximum value between the current value of `mx` and the current node's value.
14 : Push Left Child
if(root->left) q.push(root->left);
Pushes the left child of the current node to the queue if it exists.
15 : Push Right Child
if(root->right) q.push(root->right);
Pushes the right child of the current node to the queue if it exists.
16 : Store Max Value for Level
res.push_back(mx);
Stores the largest value of the current level (`mx`) into the result vector.
17 : Return Result
return res;
Returns the result vector containing the largest value from each level of the binary tree.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) where n is the number of nodes in the tree because we process each node exactly once.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) in the worst case due to the queue storing nodes at each level. In the best case (a skewed tree), the space complexity is O(1).
